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;
int a[300004];
int rnk[300004];
int ans[300004];
int bit[300004];
void update(int idx, int val) {
  while (idx <= n) {
    bit[idx] += val;
    idx += idx & -idx;
  }
}
int query(int idx) {
  int ans = 0;
  while (idx) {
    ans += bit[idx];
    idx -= idx & -idx;
  }
  return ans;
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    rnk[a[i]] = i;
  }
  update(1, 1);
  ans[1] = 1;
  for (int i = 2; i <= n; i++) {
    if (rnk[i] < rnk[i - 1]) update(i, 1), ans[i] = 1;
  }
  int q;
  cin >> q;
  int t, x, y;
  bool dirty = true;
  while (q--) {
    cin >> t >> x >> y;
    int a1 = query(y);
    int a2 = a1 - query(x - 1) + (rnk[x] > rnk[x - 1] ? 1 : 0);
    if (t == 1) {
      if (x == 1)
        cout << a1 << "\n";
      else
        cout << a2 << "\n";
    } else {
      if (1) {
        if (ans[a[x]] == 1) {
          update(a[x], -1);
          ans[a[x]] = 0;
        }
        if (ans[a[y]] == 1) {
          update(a[y], -1);
          ans[a[y]] = 0;
        }
        if (ans[a[x] + 1] == 1) {
          update(a[x] + 1, -1);
          ans[a[x] + 1] = 0;
        }
        if (ans[a[y] + 1] == 1) {
          update(a[y] + 1, -1);
          ans[a[y] + 1] = 0;
        }
      }
      int tmp = a[x];
      a[x] = a[y];
      a[y] = tmp;
      rnk[a[y]] = y;
      rnk[a[x]] = x;
      if (a[y] < a[x]) {
        int tx, ty;
        tx = x;
        ty = y;
        y = a[ty];
        x = a[tx];
      } else {
        int tx, ty;
        tx = x;
        ty = y;
        x = a[ty];
        y = a[tx];
      }
      if (rnk[x] < rnk[x - 1] && x != y + 1) {
        ans[x] = 1;
        update(x, 1);
      }
      if (rnk[x + 1] < rnk[x]) {
        ans[x + 1] = 1;
        update(x + 1, 1);
      }
      if (rnk[y + 1] < rnk[y]) {
        ans[y + 1] = 1;
        update(y + 1, 1);
      }
      if (y == 1 || rnk[y] < rnk[y - 1]) {
        ans[y] = 1;
        update(y, 1);
      }
    }
  }
  return 0;
}
