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
#include <bits/stdc++.h>
using namespace std;
int table[333888];
int n, q, pos[333888], a[333888], val[333888];
int main() {
  memset(pos, 0, sizeof(pos));
  ;
  memset(a, 0, sizeof(a));
  ;
  memset(val, 0, sizeof(val));
  ;
  scanf("%d", &n);
  for (int i = 0; i < (n); i++) {
    scanf("%d", a + (i + 1));
    pos[a[i + 1]] = i + 1;
  }
  a[n + 1] = n + 1;
  pos[n + 1] = n + 1;
  {
    memset(table, 0, sizeof(table));
    ;
  };
  for (int i = (1); i < (n + 1); i++)
    if (pos[i] < pos[i - 1]) {
      val[i] = 1;
      {
        for (int x = (i); x < 333888; x += x & -x) table[x] += (1);
      };
    }
  scanf("%d", &q);
  for (int iq = 0; iq < (q); iq++) {
    int t, l, r;
    scanf("%d%d%d", &t, &l, &r);
    if (t == 1) {
      int up, down;
      {
        up = 0;
        for (int x = (r); x > 0; x -= x & -x) up += table[x];
      };
      {
        down = 0;
        for (int x = (l); x > 0; x -= x & -x) down += table[x];
      };
      int ret = 1 + up - down;
      printf("%d\n", ret);
    } else if (l != r) {
      int vl = a[l], vr = a[r];
      swap(pos[a[l]], pos[a[r]]);
      swap(a[l], a[r]);
      if (val[vl] != 0) {
        {
          for (int x = (vl); x < 333888; x += x & -x) table[x] += (-val[vl]);
        };
        val[vl] = 0;
      }
      if (val[vl + 1] != 0) {
        {
          for (int x = (vl + 1); x < 333888; x += x & -x)
            table[x] += (-val[vl + 1]);
        };
        val[vl + 1] = 0;
      }
      if (val[vr] != 0) {
        {
          for (int x = (vr); x < 333888; x += x & -x) table[x] += (-val[vr]);
        };
        val[vr] = 0;
      }
      if (val[vr + 1] != 0) {
        {
          for (int x = (vr + 1); x < 333888; x += x & -x)
            table[x] += (-val[vr + 1]);
        };
        val[vr + 1] = 0;
      }
      if (pos[vl] < pos[vl - 1] && val[vl] == 0) {
        val[vl] = 1;
        {
          for (int x = (vl); x < 333888; x += x & -x) table[x] += (1);
        };
      }
      if (pos[vl + 1] < pos[vl] && val[vl + 1] == 0) {
        val[vl + 1] = 1;
        {
          for (int x = (vl + 1); x < 333888; x += x & -x) table[x] += (1);
        };
      }
      if (pos[vr] < pos[vr - 1] && val[vr] == 0) {
        val[vr] = 1;
        {
          for (int x = (vr); x < 333888; x += x & -x) table[x] += (1);
        };
      }
      if (pos[vr + 1] < pos[vr] && val[vr + 1] == 0) {
        val[vr + 1] = 1;
        {
          for (int x = (vr + 1); x < 333888; x += x & -x) table[x] += (1);
        };
      }
    }
  }
  return 0;
}
