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 seg[1 << 20] = {};
void update(int a, int x) {
  seg[a + (1 << 19) - 1] = x;
  a += (1 << 19) - 1;
  while (a > 0) {
    a = (a - 1) / 2;
    seg[a] = seg[a * 2 + 1] + seg[a * 2 + 2];
  }
}
int get(int a, int b, int k, int l, int r) {
  if (a > r || b < l) return 0;
  if (a <= l && r <= b) return seg[k];
  int le = get(a, b, k * 2 + 1, l, l + (r - l + 1) / 2 - 1);
  int ri = get(a, b, k * 2 + 2, l + (r - l + 1) / 2, r);
  return le + ri;
}
int main() {
  int n;
  vector<int> num(n);
  vector<int> prev(n);
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> num[i];
  }
  for (int i = 0; i < n; i++) {
    prev[num[i]] = i;
  }
  for (int i = 1; i < n; i++) {
    if (prev[i] > prev[i + 1]) {
      update(i, 1);
    } else {
      update(i, 0);
    }
  }
  int m;
  cin >> m;
  for (int i = 0; i < m; i++) {
    int t;
    cin >> t;
    if (t == 1) {
      int a, b;
      cin >> a >> b;
      int ret = get(a, b - 1, 0, 0, (1 << 19) - 1);
      cout << ret + 1 << endl;
    } else {
      int a, b;
      cin >> a >> b;
      int g = num[a - 1];
      int h = num[b - 1];
      swap(prev[g], prev[h]);
      swap(num[a - 1], num[b - 1]);
      if (g > 1) {
        if (prev[g - 1] > prev[g]) {
          update(g - 1, 1);
        } else {
          update(g - 1, 0);
        }
      }
      if (g < n) {
        if (prev[g] > prev[g + 1]) {
          update(g, 1);
        } else {
          update(g, 0);
        }
      }
      if (h > 1) {
        if (prev[h - 1] > prev[h]) {
          update(h - 1, 1);
        } else {
          update(h - 1, 0);
        }
      }
      if (h < n) {
        if (prev[h] > prev[h + 1]) {
          update(h, 1);
        } else {
          update(h, 0);
        }
      }
    }
  }
}
