#include <bits/stdc++.h>
using namespace std;
const int N = 300005;
const int inf = 2000000005;
struct Node {
  int left, right, sum;
} L[N << 2];
int n, q, a[N], b[N], pos[N] = {0};
bool cmp(int i, int j) { return a[i] < a[j]; }
void bulid(int step, int l, int r) {
  L[step].left = l;
  L[step].right = r;
  if (l == r) {
    L[step].sum = pos[l];
    return;
  }
  int m = (l + r) >> 1;
  bulid(step << 1, l, m);
  bulid(step << 1 | 1, m + 1, r);
  L[step].sum = L[step << 1].sum + L[step << 1 | 1].sum;
}
void update(int step, int pos, int val) {
  if (pos < 1 || pos > n) return;
  if (L[step].left == pos && pos == L[step].right) {
    L[step].sum = val;
    return;
  }
  int m = (L[step].left + L[step].right) >> 1;
  if (pos <= m)
    update(step << 1, pos, val);
  else
    update(step << 1 | 1, pos, val);
  L[step].sum = L[step << 1].sum + L[step << 1 | 1].sum;
}
int query(int step, int l, int r) {
  if (L[step].left == l && L[step].right == r) return L[step].sum;
  int m = (L[step].left + L[step].right) >> 1;
  if (r <= m)
    return query(step << 1, l, r);
  else if (l > m)
    return query(step << 1 | 1, l, r);
  else
    return query(step << 1, l, m) + query(step << 1 | 1, m + 1, r);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    b[i] = i;
  }
  sort(b + 1, b + 1 + n, cmp);
  b[0] = n + 1;
  b[n + 1] = n + 1;
  for (int i = n - 1; i >= 1; i--) {
    if (b[i] > b[i + 1]) pos[i + 1] = 1;
  }
  pos[1] = 1;
  bulid(1, 1, n);
  scanf("%d", &q);
  while (q--) {
    int k, x, y;
    scanf("%d%d%d", &k, &x, &y);
    if (k == 2) {
      int u = a[x], v = a[y];
      if (u > v) swap(u, v);
      if (v == u + 1) {
        if (b[u] > b[v]) {
          update(1, v, 0);
          if (b[v] < b[u - 1])
            update(1, u, 1);
          else
            update(1, u, 0);
          if (b[u] > b[v + 1])
            update(1, v + 1, 1);
          else
            update(1, v + 1, 0);
        } else {
          update(1, v, 1);
          if (b[v] < b[u - 1])
            update(1, u, 1);
          else
            update(1, u, 0);
          if (b[u] > b[v + 1])
            update(1, v + 1, 1);
          else
            update(1, v + 1, 0);
        }
      } else {
        if (b[u] > b[v]) {
          if (b[v] < b[u - 1])
            update(1, u, 1);
          else
            update(1, u, 0);
          if (b[v] > b[u + 1])
            update(1, u + 1, 1);
          else
            update(1, u + 1, 0);
          if (b[u] > b[v - 1])
            update(1, v, 0);
          else
            update(1, v, 1);
          if (b[u] > b[v + 1])
            update(1, v + 1, 1);
          else
            update(1, v + 1, 0);
        } else {
          if (b[u] < b[v - 1])
            update(1, v, 1);
          else
            update(1, v, 0);
          if (b[u] > b[v + 1])
            update(1, v + 1, 1);
          else
            update(1, v + 1, 0);
          if (b[v] > b[u - 1])
            update(1, u, 0);
          else
            update(1, u, 1);
          if (b[v] > b[u + 1])
            update(1, u + 1, 1);
          else
            update(1, u + 1, 0);
        }
      }
      swap(a[x], a[y]);
      swap(b[u], b[v]);
    } else {
      int ret = query(1, x, y) + (query(1, x, x) == 0 ? 1 : 0);
      printf("%d\n", ret);
    }
  }
  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;
const int N = 300005;
const int inf = 2000000005;
struct Node {
  int left, right, sum;
} L[N << 2];
int n, q, a[N], b[N], pos[N] = {0};
bool cmp(int i, int j) { return a[i] < a[j]; }
void bulid(int step, int l, int r) {
  L[step].left = l;
  L[step].right = r;
  if (l == r) {
    L[step].sum = pos[l];
    return;
  }
  int m = (l + r) >> 1;
  bulid(step << 1, l, m);
  bulid(step << 1 | 1, m + 1, r);
  L[step].sum = L[step << 1].sum + L[step << 1 | 1].sum;
}
void update(int step, int pos, int val) {
  if (pos < 1 || pos > n) return;
  if (L[step].left == pos && pos == L[step].right) {
    L[step].sum = val;
    return;
  }
  int m = (L[step].left + L[step].right) >> 1;
  if (pos <= m)
    update(step << 1, pos, val);
  else
    update(step << 1 | 1, pos, val);
  L[step].sum = L[step << 1].sum + L[step << 1 | 1].sum;
}
int query(int step, int l, int r) {
  if (L[step].left == l && L[step].right == r) return L[step].sum;
  int m = (L[step].left + L[step].right) >> 1;
  if (r <= m)
    return query(step << 1, l, r);
  else if (l > m)
    return query(step << 1 | 1, l, r);
  else
    return query(step << 1, l, m) + query(step << 1 | 1, m + 1, r);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    b[i] = i;
  }
  sort(b + 1, b + 1 + n, cmp);
  b[0] = n + 1;
  b[n + 1] = n + 1;
  for (int i = n - 1; i >= 1; i--) {
    if (b[i] > b[i + 1]) pos[i + 1] = 1;
  }
  pos[1] = 1;
  bulid(1, 1, n);
  scanf("%d", &q);
  while (q--) {
    int k, x, y;
    scanf("%d%d%d", &k, &x, &y);
    if (k == 2) {
      int u = a[x], v = a[y];
      if (u > v) swap(u, v);
      if (v == u + 1) {
        if (b[u] > b[v]) {
          update(1, v, 0);
          if (b[v] < b[u - 1])
            update(1, u, 1);
          else
            update(1, u, 0);
          if (b[u] > b[v + 1])
            update(1, v + 1, 1);
          else
            update(1, v + 1, 0);
        } else {
          update(1, v, 1);
          if (b[v] < b[u - 1])
            update(1, u, 1);
          else
            update(1, u, 0);
          if (b[u] > b[v + 1])
            update(1, v + 1, 1);
          else
            update(1, v + 1, 0);
        }
      } else {
        if (b[u] > b[v]) {
          if (b[v] < b[u - 1])
            update(1, u, 1);
          else
            update(1, u, 0);
          if (b[v] > b[u + 1])
            update(1, u + 1, 1);
          else
            update(1, u + 1, 0);
          if (b[u] > b[v - 1])
            update(1, v, 0);
          else
            update(1, v, 1);
          if (b[u] > b[v + 1])
            update(1, v + 1, 1);
          else
            update(1, v + 1, 0);
        } else {
          if (b[u] < b[v - 1])
            update(1, v, 1);
          else
            update(1, v, 0);
          if (b[u] > b[v + 1])
            update(1, v + 1, 1);
          else
            update(1, v + 1, 0);
          if (b[v] > b[u - 1])
            update(1, u, 0);
          else
            update(1, u, 1);
          if (b[v] > b[u + 1])
            update(1, u + 1, 1);
          else
            update(1, u + 1, 0);
        }
      }
      swap(a[x], a[y]);
      swap(b[u], b[v]);
    } else {
      int ret = query(1, x, y) + (query(1, x, x) == 0 ? 1 : 0);
      printf("%d\n", ret);
    }
  }
  return 0;
}

```