#include <bits/stdc++.h>
using namespace std;
int arr[300001];
int where[300001];
int tree[300005];
int n;
bool isBad(int val) {
  if (val == 1) return false;
  if (where[val - 1] > where[val])
    return true;
  else
    return false;
}
void add(int idx, int v) {
  while (idx <= n) {
    tree[idx] += v;
    idx += (idx & (-idx));
  }
}
int ask(int idx) {
  int ans = 0;
  while (idx > 0) {
    ans += tree[idx];
    idx -= (idx & (-idx));
  }
  return ans;
}
int query(int a, int b) {
  if (a == 1)
    return ask(b);
  else
    return ask(b) - ask(a - 1);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &arr[i]);
    where[arr[i]] = i;
  }
  for (int i = 1; i <= n; i++) {
    if (isBad(i)) {
      add(i, 1);
    }
  }
  int t;
  scanf("%d", &t);
  while (t--) {
    int type;
    int x, y;
    scanf("%d %d %d", &type, &x, &y);
    if (type == 1) {
      int ans = 1;
      ans += query(x, y);
      if (isBad(x)) {
        ans--;
      }
      printf("%d\n", ans);
    } else {
      vector<int> vvalu;
      vvalu.push_back(arr[x]);
      vvalu.push_back(arr[y]);
      if (arr[x] != n) vvalu.push_back(arr[x] + 1);
      if (arr[y] != n) vvalu.push_back(arr[y] + 1);
      while (1) {
        sort(vvalu.begin(), vvalu.end());
        bool ch = false;
        for (int i = 1; i < vvalu.size(); i++) {
          if (vvalu[i] == vvalu[i - 1]) {
            swap(vvalu[i], vvalu[vvalu.size() - 1]);
            vvalu.pop_back();
            ch = true;
            break;
          }
        }
        if (!ch) break;
      }
      int v1 = arr[x];
      int v2 = arr[y];
      for (int p = 0; p < vvalu.size(); p++) {
        if (isBad(vvalu[p])) {
          add(vvalu[p], -1);
        }
      }
      swap(arr[x], arr[y]);
      swap(where[v1], where[v2]);
      for (int p = 0; p < vvalu.size(); p++) {
        if (isBad(vvalu[p])) {
          add(vvalu[p], 1);
        }
      }
    }
  }
  return 0;
}

### Prompt

In CPP, your task is to solve 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>
using namespace std;
int arr[300001];
int where[300001];
int tree[300005];
int n;
bool isBad(int val) {
  if (val == 1) return false;
  if (where[val - 1] > where[val])
    return true;
  else
    return false;
}
void add(int idx, int v) {
  while (idx <= n) {
    tree[idx] += v;
    idx += (idx & (-idx));
  }
}
int ask(int idx) {
  int ans = 0;
  while (idx > 0) {
    ans += tree[idx];
    idx -= (idx & (-idx));
  }
  return ans;
}
int query(int a, int b) {
  if (a == 1)
    return ask(b);
  else
    return ask(b) - ask(a - 1);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &arr[i]);
    where[arr[i]] = i;
  }
  for (int i = 1; i <= n; i++) {
    if (isBad(i)) {
      add(i, 1);
    }
  }
  int t;
  scanf("%d", &t);
  while (t--) {
    int type;
    int x, y;
    scanf("%d %d %d", &type, &x, &y);
    if (type == 1) {
      int ans = 1;
      ans += query(x, y);
      if (isBad(x)) {
        ans--;
      }
      printf("%d\n", ans);
    } else {
      vector<int> vvalu;
      vvalu.push_back(arr[x]);
      vvalu.push_back(arr[y]);
      if (arr[x] != n) vvalu.push_back(arr[x] + 1);
      if (arr[y] != n) vvalu.push_back(arr[y] + 1);
      while (1) {
        sort(vvalu.begin(), vvalu.end());
        bool ch = false;
        for (int i = 1; i < vvalu.size(); i++) {
          if (vvalu[i] == vvalu[i - 1]) {
            swap(vvalu[i], vvalu[vvalu.size() - 1]);
            vvalu.pop_back();
            ch = true;
            break;
          }
        }
        if (!ch) break;
      }
      int v1 = arr[x];
      int v2 = arr[y];
      for (int p = 0; p < vvalu.size(); p++) {
        if (isBad(vvalu[p])) {
          add(vvalu[p], -1);
        }
      }
      swap(arr[x], arr[y]);
      swap(where[v1], where[v2]);
      for (int p = 0; p < vvalu.size(); p++) {
        if (isBad(vvalu[p])) {
          add(vvalu[p], 1);
        }
      }
    }
  }
  return 0;
}

```