There is an array $a$ of $2^{30}$ integers, indexed from $0$ to $2^{30}-1$. Initially, you know that $0 \leq a_i < 2^{30}$ ($0 \leq i < 2^{30}$), but you do not know any of the values. Your task is to process queries of two types:
- 1 l r x: You are informed that the bitwise xor of the subarray $[l, r]$ (ends inclusive) is equal to $x$. That is, $a_l \oplus a_{l+1} \oplus \ldots \oplus a_{r-1} \oplus a_r = x$, where $\oplus$ is the bitwise xor operator. In some cases, the received update contradicts past updates. In this case, you should ignore the contradicting update (the current update).
- 2 l r: You are asked to output the bitwise xor of the subarray $[l, r]$ (ends inclusive). If it is still impossible to know this value, considering all past updates, then output $-1$.
Note that the queries are encoded. That is, you need to write an online solution.
Input
The first line contains a single integer $q$ ($1 \leq q \leq 2 \cdot 10^5$) — the number of queries.
Each of the next $q$ lines describes a query. It contains one integer $t$ ($1 \leq t \leq 2$) — the type of query.
The given queries will be encoded in the following way: let $last$ be the answer to the last query of the second type that you have answered (initially, $last = 0$). If the last answer was $-1$, set $last = 1$.
- If $t = 1$, three integers follow, $l’$, $r’$, and $x’$ ($0 \leq l’, r’, x’ < 2^{30}$), meaning that you got an update. First, do the following:$l = l’ \oplus last$, $r = r’ \oplus last$, $x = x’ \oplus last$and, if $l > r$, swap $l$ and $r$.This means you got an update that the bitwise xor of the subarray $[l, r]$ is equal to $x$ (notice that you need to ignore updates that contradict previous updates).
- If $t = 2$, two integers follow, $l’$ and $r’$ ($0 \leq l’, r’ < 2^{30}$), meaning that you got a query. First, do the following:$l = l’ \oplus last$, $r = r’ \oplus last$and, if $l > r$, swap $l$ and $r$.For the given query, you need to print the bitwise xor of the subarray $[l, r]$. If it is impossible to know, print $-1$. Don’t forget to change the value of $last$.
It is guaranteed there will be at least one query of the second type.
Output
After every query of the second type, output the bitwise xor of the given subarray or $-1$ if it is still impossible to know.
Examples
input
12
2 1 2
2 1 1073741822
1 0 3 4
2 0 0
2 3 3
2 0 3
1 6 7 3
2 4 4
1 0 2 1
2 0 0
2 4 4
2 0 0
output
-1
-1
-1
-1
5
-1
6
3
5
input
4
1 5 5 9
1 6 6 5
1 6 5 10
2 6 5
output
12
Note
In the first example, the real queries (without being encoded) are:
- 12
- 2 1 2
- 2 0 1073741823
- 1 1 2 5
- 2 1 1
- 2 2 2
- 2 1 2
- 1 2 3 6
- 2 1 1
- 1 1 3 0
- 2 1 1
- 2 2 2
- 2 3 3
- The answers for the first two queries are $-1$ because we don’t have any such information on the array initially.
- The first update tells us $a_1 \oplus a_2 = 5$. Note that we still can’t be certain about the values $a_1$ or $a_2$ independently (for example, it could be that $a_1 = 1, a_2 = 4$, and also $a_1 = 3, a_2 = 6$).
- After we receive all three updates, we have enough information to deduce $a_1, a_2, a_3$ independently.
In the second example, notice that after the first two updates we already know that $a_5 \oplus a_6 = 12$, so the third update is contradicting, and we ignore it.
Solution:
#include <bits/stdc++.h> using namespace std; int main() { ios::sync_with_stdio(false); cin.tie(0); int tt; cin >> tt; map<int,int> mp; vector<int> p; vector<int> xr; int n = 0; auto get_id = [&](int v) { auto it = mp.find(v); if (it != mp.end()) { return it->second; } mp[v] = n; p.push_back(n); xr.push_back(0); n++; return n - 1; }; function<int(int)> find_set = [&](int x) { if (x != p[x]) { int res = find_set(p[x]); xr[x] ^= xr[p[x]]; p[x] = res; } return p[x]; }; int ans = 0; while (tt--) { int op; cin >> op; if (op == 1) { int x, y, z; cin >> x >> y >> z; x ^= ans; y ^= ans; z ^= ans; if (x > y) swap(x, y); // cerr << "query 1 " << x << " " << y << " " << z << '\n'; y++; x = get_id(x); y = get_id(y); int xx = find_set(x); int yy = find_set(y); if (xx != yy) { p[xx] = yy; xr[xx] = xr[x] ^ xr[y] ^ z; } } else { int x, y; cin >> x >> y; x ^= ans; y ^= ans; if (x > y) swap(x, y); // cerr << "query 2 " << x << " " << y << '\n'; y++; x = get_id(x); y = get_id(y); int xx = find_set(x); int yy = find_set(y); if (xx == yy) { ans = xr[x] ^ xr[y]; } else { ans = -1; } cout << ans << '\n'; ans = abs(ans); } } return 0; }