#include <bits/stdc++.h>
using namespace std;
const int inf = int(1e9);
const double eps = 1e-9;
const double pi = 4 * atan(double(1));
const int N = int(3e5);
int maxv;
int a[N + 10], p[N + 10];
int rsq[4 * N + 10];
inline int get(int l, int r) {
  int res = 0;
  l += maxv;
  r += maxv;
  while (l <= r) {
    if (l % 2 != 0) {
      res += rsq[l];
    }
    l = (l + 1) / 2;
    if (r % 2 == 0) {
      res += rsq[r];
    }
    r = (r - 1) / 2;
  }
  return res;
}
inline void update(int a, int b) {
  a += maxv;
  rsq[a] = b;
  while (a > 1) {
    a /= 2;
    rsq[a] = rsq[a * 2] + rsq[a * 2 + 1];
  }
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    a[i]--;
    p[a[i]] = i;
  }
  maxv = 1;
  while (maxv < n) {
    maxv *= 2;
  }
  for (int i = 0; i < n - 1; i++) {
    if (p[i] > p[i + 1]) {
      rsq[i + maxv] = 1;
    }
  }
  for (int i = maxv - 1; i > 0; i--) {
    rsq[i] = rsq[i * 2] + rsq[i * 2 + 1];
  }
  int m;
  scanf("%d", &m);
  for (int i = 0; i < m; i++) {
    int t, x, y;
    scanf("%d %d %d", &t, &x, &y);
    x--;
    y--;
    if (t == 1) {
      int ans = get(x, y - 1) + 1;
      printf("%d\n", ans);
      continue;
    }
    p[a[x]] = y;
    p[a[y]] = x;
    swap(a[x], a[y]);
    if (a[x] > 0) {
      update(a[x] - 1, p[a[x] - 1] > p[a[x]]);
    }
    update(a[x], (a[x] + 1 < n && p[a[x]] > p[a[x] + 1]));
    if (a[y] > 0) {
      update(a[y] - 1, p[a[y] - 1] > p[a[y]]);
    }
    update(a[y], (a[y] + 1 < n && p[a[y]] > p[a[y] + 1]));
  }
  return 0;
}

### Prompt

Your task is to create a cpp solution to 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;
const int inf = int(1e9);
const double eps = 1e-9;
const double pi = 4 * atan(double(1));
const int N = int(3e5);
int maxv;
int a[N + 10], p[N + 10];
int rsq[4 * N + 10];
inline int get(int l, int r) {
  int res = 0;
  l += maxv;
  r += maxv;
  while (l <= r) {
    if (l % 2 != 0) {
      res += rsq[l];
    }
    l = (l + 1) / 2;
    if (r % 2 == 0) {
      res += rsq[r];
    }
    r = (r - 1) / 2;
  }
  return res;
}
inline void update(int a, int b) {
  a += maxv;
  rsq[a] = b;
  while (a > 1) {
    a /= 2;
    rsq[a] = rsq[a * 2] + rsq[a * 2 + 1];
  }
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    a[i]--;
    p[a[i]] = i;
  }
  maxv = 1;
  while (maxv < n) {
    maxv *= 2;
  }
  for (int i = 0; i < n - 1; i++) {
    if (p[i] > p[i + 1]) {
      rsq[i + maxv] = 1;
    }
  }
  for (int i = maxv - 1; i > 0; i--) {
    rsq[i] = rsq[i * 2] + rsq[i * 2 + 1];
  }
  int m;
  scanf("%d", &m);
  for (int i = 0; i < m; i++) {
    int t, x, y;
    scanf("%d %d %d", &t, &x, &y);
    x--;
    y--;
    if (t == 1) {
      int ans = get(x, y - 1) + 1;
      printf("%d\n", ans);
      continue;
    }
    p[a[x]] = y;
    p[a[y]] = x;
    swap(a[x], a[y]);
    if (a[x] > 0) {
      update(a[x] - 1, p[a[x] - 1] > p[a[x]]);
    }
    update(a[x], (a[x] + 1 < n && p[a[x]] > p[a[x] + 1]));
    if (a[y] > 0) {
      update(a[y] - 1, p[a[y] - 1] > p[a[y]]);
    }
    update(a[y], (a[y] + 1 < n && p[a[y]] > p[a[y] + 1]));
  }
  return 0;
}

```