#include <bits/stdc++.h>
int n, q, va, vb, vc;
int pos[300008];
int place[300008];
int bit[300008];
void add(int plc, int vl) {
  int k = 1;
  while (plc <= n) {
    if (plc & k) {
      bit[plc] += vl;
      plc += k;
    }
    k += k;
  }
}
int calc(int plc) {
  int k = 1;
  int ret = 0;
  while (plc) {
    if (plc & k) {
      ret += bit[plc];
      plc -= k;
    }
    k += k;
  }
  return ret;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &pos[i]);
    place[pos[i]] = i;
  }
  place[0] = 0;
  place[n + 1] = n + 1;
  for (int i = 0; i <= n; i++) bit[i] = 0;
  for (int i = 1; i <= n; i++)
    if (place[i] < place[i - 1]) add(i, 1);
  scanf("%d", &q);
  for (int qry = 0; qry < q; qry++) {
    scanf("%d%d%d", &va, &vb, &vc);
    if (va == 1) {
      int ret = 1;
      int vl1 = calc(vc);
      int vl2 = calc(vb);
      printf("%d\n", vl1 - vl2 + 1);
    } else {
      int v1 = pos[vb];
      int v2 = pos[vc];
      int bef1 = 0;
      if (place[v1] < place[v1 - 1]) bef1 = 1;
      int bef2 = 0;
      if (place[v2] < place[v2 - 1]) bef2 = 1;
      int bef3 = 0;
      if (place[v1 + 1] < place[v1]) bef3 = 1;
      int bef4 = 0;
      if (place[v2 + 1] < place[v2]) bef4 = 1;
      place[v1] = vc;
      place[v2] = vb;
      pos[vb] = v2;
      pos[vc] = v1;
      if (place[v1] < place[v1 - 1]) {
        if (bef1 == 0) {
          add(v1, 1);
        }
      } else {
        if (bef1 == 1) {
          add(v1, -1);
        }
      }
      if (place[v2] < place[v2 - 1]) {
        if (bef2 == 0) {
          add(v2, 1);
        }
      } else {
        if (bef2 == 1) {
          add(v2, -1);
        }
      }
      if (v1 > v2 || v2 - v1 > 1) {
        if (place[v1 + 1] < place[v1]) {
          if (bef3 == 0) {
            add(v1 + 1, 1);
          }
        } else {
          if (bef3 == 1) {
            add(v1 + 1, -1);
          }
        }
      }
      if (v1 < v2 || v1 - v2 > 1) {
        if (place[v2 + 1] < place[v2]) {
          if (bef4 == 0) {
            add(v2 + 1, 1);
          }
        } else {
          if (bef4 == 1) {
            add(v2 + 1, -1);
          }
        }
      }
    }
  }
  return 0;
}

### Prompt

Please formulate a Cpp solution to the following problem:

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>
int n, q, va, vb, vc;
int pos[300008];
int place[300008];
int bit[300008];
void add(int plc, int vl) {
  int k = 1;
  while (plc <= n) {
    if (plc & k) {
      bit[plc] += vl;
      plc += k;
    }
    k += k;
  }
}
int calc(int plc) {
  int k = 1;
  int ret = 0;
  while (plc) {
    if (plc & k) {
      ret += bit[plc];
      plc -= k;
    }
    k += k;
  }
  return ret;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &pos[i]);
    place[pos[i]] = i;
  }
  place[0] = 0;
  place[n + 1] = n + 1;
  for (int i = 0; i <= n; i++) bit[i] = 0;
  for (int i = 1; i <= n; i++)
    if (place[i] < place[i - 1]) add(i, 1);
  scanf("%d", &q);
  for (int qry = 0; qry < q; qry++) {
    scanf("%d%d%d", &va, &vb, &vc);
    if (va == 1) {
      int ret = 1;
      int vl1 = calc(vc);
      int vl2 = calc(vb);
      printf("%d\n", vl1 - vl2 + 1);
    } else {
      int v1 = pos[vb];
      int v2 = pos[vc];
      int bef1 = 0;
      if (place[v1] < place[v1 - 1]) bef1 = 1;
      int bef2 = 0;
      if (place[v2] < place[v2 - 1]) bef2 = 1;
      int bef3 = 0;
      if (place[v1 + 1] < place[v1]) bef3 = 1;
      int bef4 = 0;
      if (place[v2 + 1] < place[v2]) bef4 = 1;
      place[v1] = vc;
      place[v2] = vb;
      pos[vb] = v2;
      pos[vc] = v1;
      if (place[v1] < place[v1 - 1]) {
        if (bef1 == 0) {
          add(v1, 1);
        }
      } else {
        if (bef1 == 1) {
          add(v1, -1);
        }
      }
      if (place[v2] < place[v2 - 1]) {
        if (bef2 == 0) {
          add(v2, 1);
        }
      } else {
        if (bef2 == 1) {
          add(v2, -1);
        }
      }
      if (v1 > v2 || v2 - v1 > 1) {
        if (place[v1 + 1] < place[v1]) {
          if (bef3 == 0) {
            add(v1 + 1, 1);
          }
        } else {
          if (bef3 == 1) {
            add(v1 + 1, -1);
          }
        }
      }
      if (v1 < v2 || v1 - v2 > 1) {
        if (place[v2 + 1] < place[v2]) {
          if (bef4 == 0) {
            add(v2 + 1, 1);
          }
        } else {
          if (bef4 == 1) {
            add(v2 + 1, -1);
          }
        }
      }
    }
  }
  return 0;
}

```