#include <bits/stdc++.h>
using namespace std;
int g[300300], g1[300300];
int sum[300300];
int c[300300];
int lowbit(int x) { return x & (x ^ (x - 1)); }
void add(int x, int val) {
  while (x <= 300300) {
    c[x] += val;
    x += lowbit(x);
  }
}
int getsum(int x) {
  int result = 0;
  while (x > 0) {
    result += c[x];
    x -= lowbit(x);
  }
  return result;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int tmp;
    scanf("%d", &tmp);
    g[tmp] = i;
    g1[i] = tmp;
  }
  memset(c, 0, sizeof(c));
  add(1, 1);
  for (int i = 2; i <= n; i++) {
    if (g[i] < g[i - 1]) add(i, 1);
  }
  int q;
  scanf("%d", &q);
  for (int i = 0; i < q; i++) {
    int sel, x, y;
    scanf("%d%d%d", &sel, &x, &y);
    if (sel == 2) {
      int tx = x;
      int ty = y;
      x = g1[x];
      y = g1[y];
      if (y < x) swap(x, y);
      swap(g1[tx], g1[ty]);
      if (y - x == 1) {
        if (g[x] > g[x - 1]) {
          if (g[y] < g[x - 1]) add(x, 1);
        } else {
          if (g[y] > g[x - 1]) add(x, -1);
        }
        if (g[y] < g[y + 1]) {
          if (g[x] > g[y + 1]) add(y + 1, 1);
        } else if (g[x] < g[y + 1])
          add(y + 1, -1);
        if (g[x] < g[y]) {
          add(y, 1);
        } else
          add(y, -1);
        swap(g[x], g[y]);
      } else {
        if (g[x] > g[x - 1]) {
          if (g[y] < g[x - 1]) add(x, 1);
        } else {
          if (g[y] > g[x - 1]) add(x, -1);
        }
        if (g[x] < g[x + 1]) {
          if (g[y] > g[x + 1]) add(x + 1, 1);
        } else if (g[y] < g[x + 1])
          add(x + 1, -1);
        if (g[y] > g[y - 1]) {
          if (g[x] < g[y - 1]) add(y, 1);
        } else {
          if (g[x] > g[y - 1]) add(y, -1);
        }
        if (g[y] < g[y + 1]) {
          if (g[x] > g[y + 1]) add(y + 1, 1);
        } else if (g[x] < g[y + 1])
          add(y + 1, -1);
        swap(g[x], g[y]);
      }
    } else {
      printf("%d\n", getsum(y) - getsum(x) + 1);
    }
  }
  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 g[300300], g1[300300];
int sum[300300];
int c[300300];
int lowbit(int x) { return x & (x ^ (x - 1)); }
void add(int x, int val) {
  while (x <= 300300) {
    c[x] += val;
    x += lowbit(x);
  }
}
int getsum(int x) {
  int result = 0;
  while (x > 0) {
    result += c[x];
    x -= lowbit(x);
  }
  return result;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int tmp;
    scanf("%d", &tmp);
    g[tmp] = i;
    g1[i] = tmp;
  }
  memset(c, 0, sizeof(c));
  add(1, 1);
  for (int i = 2; i <= n; i++) {
    if (g[i] < g[i - 1]) add(i, 1);
  }
  int q;
  scanf("%d", &q);
  for (int i = 0; i < q; i++) {
    int sel, x, y;
    scanf("%d%d%d", &sel, &x, &y);
    if (sel == 2) {
      int tx = x;
      int ty = y;
      x = g1[x];
      y = g1[y];
      if (y < x) swap(x, y);
      swap(g1[tx], g1[ty]);
      if (y - x == 1) {
        if (g[x] > g[x - 1]) {
          if (g[y] < g[x - 1]) add(x, 1);
        } else {
          if (g[y] > g[x - 1]) add(x, -1);
        }
        if (g[y] < g[y + 1]) {
          if (g[x] > g[y + 1]) add(y + 1, 1);
        } else if (g[x] < g[y + 1])
          add(y + 1, -1);
        if (g[x] < g[y]) {
          add(y, 1);
        } else
          add(y, -1);
        swap(g[x], g[y]);
      } else {
        if (g[x] > g[x - 1]) {
          if (g[y] < g[x - 1]) add(x, 1);
        } else {
          if (g[y] > g[x - 1]) add(x, -1);
        }
        if (g[x] < g[x + 1]) {
          if (g[y] > g[x + 1]) add(x + 1, 1);
        } else if (g[y] < g[x + 1])
          add(x + 1, -1);
        if (g[y] > g[y - 1]) {
          if (g[x] < g[y - 1]) add(y, 1);
        } else {
          if (g[x] > g[y - 1]) add(y, -1);
        }
        if (g[y] < g[y + 1]) {
          if (g[x] > g[y + 1]) add(y + 1, 1);
        } else if (g[x] < g[y + 1])
          add(y + 1, -1);
        swap(g[x], g[y]);
      }
    } else {
      printf("%d\n", getsum(y) - getsum(x) + 1);
    }
  }
  return 0;
}

```