# Red Blue Tree

You are given a tree of $n$ nodes. The tree is rooted at node $1$, which is not considered as a leaf regardless of its degree.

Each leaf of the tree has one of the two colors: red or blue. Leaf node $v$ initially has color $s_{v}$.

The color of each of the internal nodes (including the root) is determined as follows.

• Let $b$ be the number of blue immediate children, and $r$ be the number of red immediate children of a given vertex.
• Then the color of this vertex is blue if and only if $b – r \ge k$, otherwise red.

Integer $k$ is a parameter that is same for all the nodes.

You need to handle the following types of queries:

• 1 v: print the color of node $v$;
• 2 v c: change the color of leaf $v$ to $c$ ($c = 0$ means red, $c = 1$ means blue);
• 3 h: update the current value of $k$ to $h$.

Input

The first line of the input consists of two integers $n$ and $k$ ($2 \le n \le 10^{5}$, $-n \le k \le n$) — the number of nodes and the initial parameter $k$.

Each of the next $n – 1$ lines contains two integers $u$ and $v$ ($1 \le u,v \le n$), denoting that there is an edge between vertices $u$ and $v$.

The next line consists of $n$ space separated integers — the initial array $s$ ($-1 \le s_i \le 1$). $s_{i} = 0$ means that the color of node $i$ is red. $s_{i} = 1$ means that the color of node $i$ is blue. $s_{i} = -1$ means that the node $i$ is not a leaf.

The next line contains an integer $q$ ($1 \le q \le 10^5$), the number of queries.

$q$ lines follow, each containing a query in one of the following queries:

• 1 v ($1 \le v \le n$): print the color of node $v$;
• 2 v c ($1 \le v \le n$, $c = 0$ or $c = 1$): change the color of leaf $v$ to $c$ ($c = 0$ means red, $c = 1$ means blue). It is guaranteed that $v$ is a leaf;
• 3 h ($-n \le h \le n$): update the current value of $k$ to $h$.

Output

For each query of the first type, print $0$ if the color of vertex $v$ is red, and $1$ otherwise.Exampleinput

5 2
1 2
1 3
2 4
2 5
-1 -1 0 1 0
9
1 1
1 2
3 -2
1 1
1 2
3 1
2 5 1
1 1
1 2


output

0
0
1
1
0
1


NoteFigures:

(i) The initial tree

(ii) The tree after the 3rd query

(iii) The tree after the 7th query

Solution:

#pragma GCC optimize("Ofast", "unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")

#include <bits/stdc++.h>

using namespace std;

template <typename A, typename B>
string to_string(pair<A, B> p);

template <typename A, typename B, typename C>
string to_string(tuple<A, B, C> p);

template <typename A, typename B, typename C, typename D>
string to_string(tuple<A, B, C, D> p);

string to_string(const string& s) {
return '"' + s + '"';
}

string to_string(const char* s) {
}

string to_string(bool b) {
return (b ? "true" : "false");
}

string to_string(vector<bool> v) {
bool first = true;
string res = "{";
for (int i = 0; i < static_cast<int>(v.size()); i++) {
if (!first) {
res += ", ";
}
first = false;
res += to_string(v[i]);
}
res += "}";
return res;
}

template <size_t N>
string to_string(bitset<N> v) {
string res = "";
for (size_t i = 0; i < N; i++) {
res += static_cast<char>('0' + v[i]);
}
return res;
}

template <typename A>
string to_string(A v) {
bool first = true;
string res = "{";
for (const auto &x : v) {
if (!first) {
res += ", ";
}
first = false;
res += to_string(x);
}
res += "}";
return res;
}

template <typename A, typename B>
string to_string(pair<A, B> p) {
return "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
}

template <typename A, typename B, typename C>
string to_string(tuple<A, B, C> p) {
return "(" + to_string(get<0>(p)) + ", " + to_string(get<1>(p)) + ", " + to_string(get<2>(p)) + ")";
}

template <typename A, typename B, typename C, typename D>
string to_string(tuple<A, B, C, D> p) {
return "(" + to_string(get<0>(p)) + ", " + to_string(get<1>(p)) + ", " + to_string(get<2>(p)) + ", " + to_string(get<3>(p)) + ")";
}

void debug_out() { cerr << endl; }

void debug_out(Head H, Tail... T) {
cerr << " " << to_string(H);
debug_out(T...);
}

#ifdef LOCAL
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#else
#define debug(...) 42
#endif

const int N = 100010;

int from_leaves[N];
int from_others[N];

int main() {
ios::sync_with_stdio(false);
cin.tie(0);
int n, k;
cin >> n >> k;
vector<vector<int>> g(n);
for (int i = 0; i < n - 1; i++) {
int x, y;
cin >> x >> y;
--x; --y;
g[x].push_back(y);
g[y].push_back(x);
}
vector<int> color(n);
for (int i = 0; i < n; i++) {
cin >> color[i];
}
vector<int> order;
vector<int> pos(n, -1);
vector<int> endd(n, -1);
vector<bool> is_leaf(n, false);
vector<int> map_to(n);
vector<int> deg(n);
vector<int> parent(n, -1);
iota(map_to.begin(), map_to.end(), 0);
function<void(int, int)> Dfs = [&](int v, int pv) {
parent[v] = pv;
deg[v] = (int) g[v].size() - (pv != -1);
if (pv != -1 && deg[v] == 1 && deg[pv] == 1) {
map_to[v] = map_to[pv];
for (int u : g[v]) {
if (u == pv) {
continue;
}
Dfs(u, v);
}
return;
}
pos[v] = (int) order.size();
order.push_back(v);
for (int u : g[v]) {
if (u == pv) {
continue;
}
Dfs(u, v);
}
endd[v] = (int) order.size() - 1;
if (deg[v] == 0) {
is_leaf[v] = true;
order.pop_back();
pos[v] = endd[v] = -1;
}
};
Dfs(0, -1);
int cnt = (int) order.size();
memset(from_leaves, 0, sizeof(int) * cnt);
memset(from_others, 0, sizeof(int) * cnt);
for (int i = 0; i < n; i++) {
if (is_leaf[i]) {
from_leaves[pos[map_to[parent[i]]]] += (color[i] == 1 ? 1 : -1);
}
}
vector<int> go_to(cnt);
for (int i = 1; i < cnt; i++) {
go_to[i] = pos[map_to[parent[order[i]]]];
}
int tt;
cin >> tt;
vector<int> op(tt), ver(tt), val(tt), res(tt, -1);
for (int i = 0; i < tt; i++) {
cin >> op[i];
if (op[i] == 1) {
cin >> ver[i];
--ver[i];
}
if (op[i] == 2) {
cin >> ver[i] >> val[i];
--ver[i];
}
if (op[i] == 3) {
cin >> val[i];
}
}
map<int, int> mp;
int beg = 0;
while (beg < tt) {
int end = beg;
while (end + 1 < tt && op[end + 1] == op[end]) {
++end;
}
if (op[beg] == 1) {
mp.clear();
for (int i = beg; i <= end; i++) {
int v = ver[i];
if (!is_leaf[v]) {
v = map_to[v];
from_others[pos[v]] = 0;
mp[pos[v] + 1] += 1;
mp[endd[v] + 1] -= 1;
}
}
vector<pair<int, int>> segs;
int balance = 0;
int start = -1;
for (auto& p : mp) {
if (p.second == 0) {
continue;
}
if (balance == 0) {
start = p.first;
}
balance += p.second;
if (balance == 0) {
segs.emplace_back(start, p.first - 1);
start = -1;
}
}
for (auto& p : segs) {
int from = p.first;
int to = p.second;
memset(from_others + from, 0, sizeof(int) * (to - from + 1));
for (int i = to; i >= from; i--) {
from_others[go_to[i]] += (from_others[i] + from_leaves[i] >= k ? 1 : -1);
}
}
for (int i = beg; i <= end; i++) {
int v = ver[i];
if (is_leaf[v]) {
res[i] = color[v];
} else {
v = map_to[v];
res[i] = (from_others[pos[v]] + from_leaves[pos[v]] >= k);
}
}
}
if (op[beg] == 2) {
for (int i = beg; i <= end; i++) {
int v = ver[i];
from_leaves[pos[map_to[parent[v]]]] -= (color[v] == 1 ? 1 : -1);
color[v] = val[i];
from_leaves[pos[map_to[parent[v]]]] += (color[v] == 1 ? 1 : -1);
}
}
if (op[beg] == 3) {
k = val[end];
}
beg = end + 1;
}
for (int i = 0; i < tt; i++) {
if (op[i] == 1) {
cout << res[i] << '\n';
}
}
return 0;
}