#include <bits/stdc++.h>
using namespace std;
const int N = 400000;
int ar[N], pos[N], val[N];
inline int lowbit(int x) { return x & (-x); }
int sum(int i) {
  int s = 0;
  for (; i > 0; s += ar[i], i -= lowbit(i))
    ;
  return s;
}
void add(int i, int v) {
  for (; i < N; ar[i] += v, i += lowbit(i))
    ;
}
int main() {
  int n, m;
  while (scanf("%d", &n) != EOF) {
    memset(ar, 0, sizeof(ar));
    for (int i = 1; i <= n; i++) {
      scanf("%d", &val[i]);
      pos[val[i]] = i;
    }
    for (int i = 1; i < n; i++) {
      if (pos[i] > pos[i + 1]) {
        add(i, 1);
      }
    }
    scanf("%d", &m);
    while (m--) {
      int op, a, b, va, vb;
      scanf("%d%d%d", &op, &a, &b);
      if (op == 1) {
        printf("%d\n", sum(b - 1) - sum(a - 1) + 1);
      } else {
        va = val[a];
        vb = val[b];
        if (va != n && pos[va + 1] < b && pos[va + 1] > a) {
          add(va, 1);
        }
        if (va != 1 && pos[va - 1] < b && pos[va - 1] > a) {
          add(va - 1, -1);
        }
        if (vb != n && pos[vb + 1] < b && pos[vb + 1] > a) {
          add(vb, -1);
        }
        if (vb != 1 && pos[vb - 1] < b && pos[vb - 1] > a) {
          add(vb - 1, 1);
        }
        if (va + 1 == vb) {
          add(va, 1);
        }
        if (vb + 1 == va) {
          add(vb, -1);
        }
        swap(val[a], val[b]);
        swap(pos[va], pos[vb]);
      }
    }
  }
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 a1, a2, ..., an 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 i1 < i2 < ... < ik, that ai1 = x, ai2 = x + 1, ..., aik - 1 = y - 1, aik = 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, p1], [p1 + 1, p2], ..., [pm + 1, y] (x ≤ p1 < p2 < ... < pm < 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 ax and ay) 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 pi xi yi, where pi is the query type (1 is to shave beavers from xi to yi, inclusive, 2 is to swap beavers on positions xi and yi). All queries meet the condition: 1 ≤ xi < yi ≤ 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 pi = 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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 400000;
int ar[N], pos[N], val[N];
inline int lowbit(int x) { return x & (-x); }
int sum(int i) {
  int s = 0;
  for (; i > 0; s += ar[i], i -= lowbit(i))
    ;
  return s;
}
void add(int i, int v) {
  for (; i < N; ar[i] += v, i += lowbit(i))
    ;
}
int main() {
  int n, m;
  while (scanf("%d", &n) != EOF) {
    memset(ar, 0, sizeof(ar));
    for (int i = 1; i <= n; i++) {
      scanf("%d", &val[i]);
      pos[val[i]] = i;
    }
    for (int i = 1; i < n; i++) {
      if (pos[i] > pos[i + 1]) {
        add(i, 1);
      }
    }
    scanf("%d", &m);
    while (m--) {
      int op, a, b, va, vb;
      scanf("%d%d%d", &op, &a, &b);
      if (op == 1) {
        printf("%d\n", sum(b - 1) - sum(a - 1) + 1);
      } else {
        va = val[a];
        vb = val[b];
        if (va != n && pos[va + 1] < b && pos[va + 1] > a) {
          add(va, 1);
        }
        if (va != 1 && pos[va - 1] < b && pos[va - 1] > a) {
          add(va - 1, -1);
        }
        if (vb != n && pos[vb + 1] < b && pos[vb + 1] > a) {
          add(vb, -1);
        }
        if (vb != 1 && pos[vb - 1] < b && pos[vb - 1] > a) {
          add(vb - 1, 1);
        }
        if (va + 1 == vb) {
          add(va, 1);
        }
        if (vb + 1 == va) {
          add(vb, -1);
        }
        swap(val[a], val[b]);
        swap(pos[va], pos[vb]);
      }
    }
  }
  return 0;
}

```