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 n, s[333333], pz[333333], have[333333], p[333333];
void modify(int x, int v) {
  while (x <= n) {
    s[x] += v;
    x = (x | (x - 1)) + 1;
  }
}
int findsum(int x) {
  int v = 0;
  while (x > 0) {
    v += s[x];
    x &= x - 1;
  }
  return v;
}
void put(int x, int v) {
  if (have[x] != v) {
    modify(x, v - have[x]);
    have[x] = v;
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int foo;
    scanf("%d", &foo);
    p[i] = foo;
    pz[foo] = i;
  }
  for (int i = 1; i <= n; i++) s[i] = have[i] = 0;
  for (int i = 1; i < n; i++)
    if (pz[i] > pz[i + 1]) put(i, 1);
  int tt;
  scanf("%d", &tt);
  while (tt--) {
    int com, x, y;
    scanf("%d %d %d", &com, &x, &y);
    if (com == 1) {
      printf("%d\n", findsum(y - 1) - findsum(x - 1) + 1);
    } else {
      int tmp = p[x];
      p[x] = p[y];
      p[y] = tmp;
      pz[p[x]] = x;
      pz[p[y]] = y;
      {
        int z = x;
        if (p[z] > 1)
          if (pz[p[z] - 1] > pz[p[z]])
            put(p[z] - 1, 1);
          else
            put(p[z] - 1, 0);
        if (p[z] < n)
          if (pz[p[z]] > pz[p[z] + 1])
            put(p[z], 1);
          else
            put(p[z], 0);
      }
      {
        int z = y;
        if (p[z] > 1)
          if (pz[p[z] - 1] > pz[p[z]])
            put(p[z] - 1, 1);
          else
            put(p[z] - 1, 0);
        if (p[z] < n)
          if (pz[p[z]] > pz[p[z] + 1])
            put(p[z], 1);
          else
            put(p[z], 0);
      }
    }
  }
  return 0;
}
