Shave Beaver!

The Smart Beaver has recently designed and built an innovative nanotechnologic all-purpose beaver mass shaving machine, “Beavershave 5000”. Beavershave 5000 can shave beavers by families! How does it work? Very easily!

There are n beavers, each of them has a unique id from 1 to n. Consider a permutation a ₁, a ₂, …, a _n of n these beavers. Beavershave 5000 needs one session to shave beavers with ids from x to y (inclusive) if and only if there are such indices i ₁ < i ₂ < … < i _k, that a _i 1 = x, a _i 2 = x + 1, …, a _i k - 1 = y - 1, a _i k = y. And that is really convenient. For example, it needs one session to shave a permutation of beavers 1, 2, 3, …, n.

If we can’t shave beavers from x to y in one session, then we can split these beavers into groups [x, p ₁], [p ₁ + 1, p ₂], …, [p _m + 1, y] (x ≤ p ₁ < p ₂ < … < p _m < y), in such a way that the machine can shave beavers in each group in one session. But then Beavershave 5000 needs m + 1 working sessions to shave beavers from x to y.

All beavers are restless and they keep trying to swap. So if we consider the problem more formally, we can consider queries of two types:

• what is the minimum number of sessions that Beavershave 5000 needs to shave beavers with ids from x to y, inclusive?
• two beavers on positions x and y (the beavers a _x and a _y) swapped.

You can assume that any beaver can be shaved any number of times.

Input

The first line contains integer n — the total number of beavers, 2 ≤ n. The second line contains n space-separated integers — the initial beaver permutation.

The third line contains integer q — the number of queries, 1 ≤ q ≤ 10⁵. The next q lines contain the queries. Each query i looks as p _i x _i y _i, where p _i is the query type (1 is to shave beavers from x _i to y _i, inclusive, 2 is to swap beavers on positions x _i and y _i). All queries meet the condition: 1 ≤ x _i < y _i ≤ n.

• to get 30 points, you need to solve the problem with constraints: n ≤ 100 (subproblem B1);
• to get 100 points, you need to solve the problem with constraints: n ≤ 3·10⁵ (subproblems B1+B2).

Note that the number of queries q is limited 1 ≤ q ≤ 10⁵ in both subproblem B1 and subproblem B2.

Output

For each query with p _i = 1, print the minimum number of Beavershave 5000 sessions.

Examples

input

5
1 3 4 2 5
6
1 1 5
1 3 4
2 2 3
1 1 5
2 1 5
1 1 5

output

2
1
3
5

Solution:

#include <iostream>
#include <iomanip>
#include <stdio.h>
#include <set>
#include <vector>
#include <map>
#include <cmath>
#include <algorithm>
#include <memory.h>
#include <string>
#include <sstream>
#include <cstdlib>
#include <ctime>
#include <cassert>

using namespace std;

int n, s[333333], pz[333333], have[333333], p[333333];

void modify(int x, int v) {
while (x <= n) {
    s[x] += v;
    x = (x | (x-1))+1;
  }
}

int findsum(int x) {
  int v = 0;
  while (x > 0) {
v += s[x];
x &= x-1;
}
return v;
}

void put(int x, int v) {
if (have[x] != v) {
modify(x, v-have[x]);
have[x] = v;
}
}

int main() {
scanf("%d", &n);
for (int i=1;i<=n;i++) {
    int foo;
    scanf("%d", &foo);
    p[i] = foo;
    pz[foo] = i;
  }
  for (int i=1;i<=n;i++) s[i] = have[i] = 0;
  for (int i=1;i<n;i++)
    if (pz[i] > pz[i+1]) put(i, 1);
int tt;
scanf("%d", &tt);
while (tt--) {
int com, x, y;
scanf("%d %d %d", &com, &x, &y);
if (com == 1) {
printf("%d\n", findsum(y-1) - findsum(x-1) + 1);
} else {
int tmp = p[x]; p[x] = p[y]; p[y] = tmp;
pz[p[x]] = x;
pz[p[y]] = y;
{
int z = x;
if (p[z] > 1)
if (pz[p[z]-1] > pz[p[z]]) put(p[z]-1, 1);
else put(p[z]-1, 0);
if (p[z] < n)
          if (pz[p[z]] > pz[p[z]+1]) put(p[z], 1);
else put(p[z], 0);
}
{
int z = y;
if (p[z] > 1)
if (pz[p[z]-1] > pz[p[z]]) put(p[z]-1, 1);
else put(p[z]-1, 0);
if (p[z] < n)
          if (pz[p[z]] > pz[p[z]+1]) put(p[z], 1);
else put(p[z], 0);
}
}
}
return 0;
}