#include <bits/stdc++.h>
using namespace std;
int sss[444444], p[444444], a[444444], n, b[444444], c[444444];
inline void upd(int x, int v) {
  for (int j = x; j < n + 11; j |= (j + 1)) sss[j] += v;
}
inline int fs(int x) {
  if (x < 0) return 0;
  int res = 0;
  for (int j = x; j >= 0; j = (j & (j + 1)) - 1) res += sss[j];
  return res;
}
inline void changel(int x, int y, int p) {
  if (!p) return;
  if (p < x && p < y) return;
  if (p > x && p > y) return;
  if (p < x && p > y)
    upd(a[p], 1);
  else if (p > x && p < y)
    upd(a[p], -1);
}
inline void changer(int x, int y, int p) {
  if (!p) return;
  if (p > x && p > y) return;
  if (p < x && p < y) return;
  if (p > x && p < y)
    upd(a[x], 1);
  else if (p < x && p > y)
    upd(a[x], -1);
}
inline void swp(int x, int y) {
  if (x > y) swp(x, y);
  if (((a[x] - a[y]) > 0 ? (a[x] - a[y]) : -(a[x] - a[y])) == 1) {
    if (a[x] < a[y]) {
      changel(x, y, p[a[x] - 1]);
      upd(a[x], 1);
      changer(y, x, p[a[y] + 1]);
    } else {
      changel(y, x, p[a[y] - 1]);
      upd(a[y], -1);
      changer(x, y, p[a[x] + 1]);
    }
  } else {
    changel(x, y, p[a[x] - 1]);
    changer(x, y, p[a[x] + 1]);
    changel(y, x, p[a[y] - 1]);
    changer(y, x, p[a[y] + 1]);
  }
  swap(a[x], a[y]);
  p[a[x]] = x;
  p[a[y]] = y;
}
int main(int argc, char* argv[]) {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]), p[a[i]] = i;
  for (int i = 1; i <= n; i++)
    if (a[i] != n && p[a[i] + 1] < i) upd(a[i], 1);
  int ttt;
  scanf("%d", &ttt);
  while (ttt--) {
    int q;
    scanf("%d", &q);
    int x, y;
    scanf("%d", &x);
    scanf("%d", &y);
    if (q == 1) {
      int ans = 0;
      if (x == y)
        ans = 0;
      else
        ans = fs(y - 1) - fs(x - 1);
      printf("%d\n", ans + 1);
    } else {
      if (x == y) continue;
      swp(x, y);
    }
  }
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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;
int sss[444444], p[444444], a[444444], n, b[444444], c[444444];
inline void upd(int x, int v) {
  for (int j = x; j < n + 11; j |= (j + 1)) sss[j] += v;
}
inline int fs(int x) {
  if (x < 0) return 0;
  int res = 0;
  for (int j = x; j >= 0; j = (j & (j + 1)) - 1) res += sss[j];
  return res;
}
inline void changel(int x, int y, int p) {
  if (!p) return;
  if (p < x && p < y) return;
  if (p > x && p > y) return;
  if (p < x && p > y)
    upd(a[p], 1);
  else if (p > x && p < y)
    upd(a[p], -1);
}
inline void changer(int x, int y, int p) {
  if (!p) return;
  if (p > x && p > y) return;
  if (p < x && p < y) return;
  if (p > x && p < y)
    upd(a[x], 1);
  else if (p < x && p > y)
    upd(a[x], -1);
}
inline void swp(int x, int y) {
  if (x > y) swp(x, y);
  if (((a[x] - a[y]) > 0 ? (a[x] - a[y]) : -(a[x] - a[y])) == 1) {
    if (a[x] < a[y]) {
      changel(x, y, p[a[x] - 1]);
      upd(a[x], 1);
      changer(y, x, p[a[y] + 1]);
    } else {
      changel(y, x, p[a[y] - 1]);
      upd(a[y], -1);
      changer(x, y, p[a[x] + 1]);
    }
  } else {
    changel(x, y, p[a[x] - 1]);
    changer(x, y, p[a[x] + 1]);
    changel(y, x, p[a[y] - 1]);
    changer(y, x, p[a[y] + 1]);
  }
  swap(a[x], a[y]);
  p[a[x]] = x;
  p[a[y]] = y;
}
int main(int argc, char* argv[]) {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]), p[a[i]] = i;
  for (int i = 1; i <= n; i++)
    if (a[i] != n && p[a[i] + 1] < i) upd(a[i], 1);
  int ttt;
  scanf("%d", &ttt);
  while (ttt--) {
    int q;
    scanf("%d", &q);
    int x, y;
    scanf("%d", &x);
    scanf("%d", &y);
    if (q == 1) {
      int ans = 0;
      if (x == y)
        ans = 0;
      else
        ans = fs(y - 1) - fs(x - 1);
      printf("%d\n", ans + 1);
    } else {
      if (x == y) continue;
      swp(x, y);
    }
  }
  return 0;
}

```