#include <bits/stdc++.h>
using namespace std;
const int maxn = 300010;
int tree[maxn];
void update(int pos, int val) {
  while (pos < maxn) tree[pos] += val, pos += pos & -pos;
}
int query(int pos) {
  int ans = 0;
  while (pos) ans += tree[pos], pos -= pos & -pos;
  return ans;
}
int pos[maxn];
int a[maxn];
int f[maxn];
int n;
int main() {
  while (~scanf("%d", &n)) {
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
    for (int i = 1; i <= n; i++) pos[a[i]] = i;
    memset(f, 0, sizeof(f));
    for (int i = 1; i < n; i++) {
      if (pos[i + 1] < pos[i]) {
        f[i] = 1;
        update(i, 1);
      }
    }
    int q;
    scanf("%d", &q);
    while (q--) {
      int type, x, y;
      scanf("%d %d %d", &type, &x, &y);
      if (type == 1) {
        int ans = query(y - 1) - query(x - 1) + 1;
        printf("%d\n", ans);
      } else {
        if (x == y) continue;
        int xx = f[a[x]];
        int yy = f[a[y]];
        pos[a[x]] = y;
        pos[a[y]] = x;
        int newx = 0, newy = 0;
        if (a[x] != n) {
          if (pos[a[x] + 1] < pos[a[x]]) newx = 1;
        }
        if (a[y] != n) {
          if (pos[a[y] + 1] < pos[a[y]]) newy = 1;
        }
        update(a[x], newx - xx);
        update(a[y], newy - yy);
        f[a[x]] = newx;
        f[a[y]] = newy;
        swap(a[x], a[y]);
        if (a[x] - 1 > 0 && a[x] - 1 != a[y]) {
          int nn = 0;
          if (pos[a[x]] < pos[a[x] - 1]) nn = 1;
          update(a[x] - 1, nn - f[a[x] - 1]);
          f[a[x] - 1] = nn;
        }
        if (a[y] - 1 > 0 && a[y] - 1 != a[x]) {
          int nn = 0;
          if (pos[a[y]] < pos[a[y] - 1]) nn = 1;
          update(a[y] - 1, nn - f[a[y] - 1]);
          f[a[y] - 1] = nn;
        }
      }
    }
  }
}

### 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 maxn = 300010;
int tree[maxn];
void update(int pos, int val) {
  while (pos < maxn) tree[pos] += val, pos += pos & -pos;
}
int query(int pos) {
  int ans = 0;
  while (pos) ans += tree[pos], pos -= pos & -pos;
  return ans;
}
int pos[maxn];
int a[maxn];
int f[maxn];
int n;
int main() {
  while (~scanf("%d", &n)) {
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
    for (int i = 1; i <= n; i++) pos[a[i]] = i;
    memset(f, 0, sizeof(f));
    for (int i = 1; i < n; i++) {
      if (pos[i + 1] < pos[i]) {
        f[i] = 1;
        update(i, 1);
      }
    }
    int q;
    scanf("%d", &q);
    while (q--) {
      int type, x, y;
      scanf("%d %d %d", &type, &x, &y);
      if (type == 1) {
        int ans = query(y - 1) - query(x - 1) + 1;
        printf("%d\n", ans);
      } else {
        if (x == y) continue;
        int xx = f[a[x]];
        int yy = f[a[y]];
        pos[a[x]] = y;
        pos[a[y]] = x;
        int newx = 0, newy = 0;
        if (a[x] != n) {
          if (pos[a[x] + 1] < pos[a[x]]) newx = 1;
        }
        if (a[y] != n) {
          if (pos[a[y] + 1] < pos[a[y]]) newy = 1;
        }
        update(a[x], newx - xx);
        update(a[y], newy - yy);
        f[a[x]] = newx;
        f[a[y]] = newy;
        swap(a[x], a[y]);
        if (a[x] - 1 > 0 && a[x] - 1 != a[y]) {
          int nn = 0;
          if (pos[a[x]] < pos[a[x] - 1]) nn = 1;
          update(a[x] - 1, nn - f[a[x] - 1]);
          f[a[x] - 1] = nn;
        }
        if (a[y] - 1 > 0 && a[y] - 1 != a[x]) {
          int nn = 0;
          if (pos[a[y]] < pos[a[y] - 1]) nn = 1;
          update(a[y] - 1, nn - f[a[y] - 1]);
          f[a[y] - 1] = nn;
        }
      }
    }
  }
}

```