#include <bits/stdc++.h>
using namespace std;
FILE *in;
FILE *out;
const int MAX = 1048576;
const int TREE = 2097152;
int n;
int a[MAX];
int ind[MAX];
int tree[TREE];
void update(int idx, int val) {
  idx += (TREE >> 1);
  while (idx) {
    tree[idx] += val;
    idx >>= 1;
  }
}
int query(int idx1, int idx2) {
  idx1 += (TREE >> 1);
  idx2 += (TREE >> 1);
  if (idx1 == idx2) return tree[idx1];
  int ret = 0;
  int flag1 = !(idx1 & 1);
  ret += tree[idx1];
  idx1 >>= 1;
  int flag2 = (idx2 & 1);
  ret += tree[idx2];
  idx2 >>= 1;
  while (idx1 != idx2) {
    if (flag1) ret += tree[(idx1 << 1) + 1];
    if (flag2) ret += tree[(idx2 << 1) + 0];
    flag1 = !(idx1 & 1);
    idx1 >>= 1;
    flag2 = (idx2 & 1);
    idx2 >>= 1;
  }
  return ret;
}
void doUpdate(int num, int oIdx, int nIdx) {
  if (oIdx < ind[num + 1] && nIdx >= ind[num + 1]) update(num, +1);
  if (oIdx > ind[num + 1] && nIdx <= ind[num + 1]) update(num, -1);
  if (oIdx < ind[num - 1] && nIdx >= ind[num - 1]) update(num - 1, -1);
  if (oIdx > ind[num - 1] && nIdx <= ind[num - 1]) update(num - 1, +1);
}
int main(void) {
  in = stdin;
  out = stdout;
  fscanf(in, "%d", &n);
  for (int i = 1; i <= n; i++) {
    fscanf(in, "%d", &a[i]);
    ind[a[i]] = i;
  }
  a[0] = 0;
  ind[0] = 0;
  a[n + 1] = n + 1;
  ind[n + 1] = n + 1;
  for (int i = 1; i <= n; i++) {
    if (ind[i + 1] < ind[i]) update(i, +1);
  }
  int numQueries;
  fscanf(in, "%d", &numQueries);
  for (int i = 0; i < numQueries; i++) {
    int t, x, y;
    fscanf(in, "%d %d %d", &t, &x, &y);
    if (t == 1) {
      fprintf(out, "%d\n", query(x, y - 1) + 1);
    } else {
      doUpdate(a[x], x, y);
      doUpdate(a[y], y, x);
      if (abs(a[x] - a[y]) == 1) {
        if (a[x] < a[y])
          update(a[x], -1);
        else
          update(a[y], +1);
      }
      swap(a[x], a[y]);
      swap(ind[a[x]], ind[a[y]]);
    }
  }
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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;
FILE *in;
FILE *out;
const int MAX = 1048576;
const int TREE = 2097152;
int n;
int a[MAX];
int ind[MAX];
int tree[TREE];
void update(int idx, int val) {
  idx += (TREE >> 1);
  while (idx) {
    tree[idx] += val;
    idx >>= 1;
  }
}
int query(int idx1, int idx2) {
  idx1 += (TREE >> 1);
  idx2 += (TREE >> 1);
  if (idx1 == idx2) return tree[idx1];
  int ret = 0;
  int flag1 = !(idx1 & 1);
  ret += tree[idx1];
  idx1 >>= 1;
  int flag2 = (idx2 & 1);
  ret += tree[idx2];
  idx2 >>= 1;
  while (idx1 != idx2) {
    if (flag1) ret += tree[(idx1 << 1) + 1];
    if (flag2) ret += tree[(idx2 << 1) + 0];
    flag1 = !(idx1 & 1);
    idx1 >>= 1;
    flag2 = (idx2 & 1);
    idx2 >>= 1;
  }
  return ret;
}
void doUpdate(int num, int oIdx, int nIdx) {
  if (oIdx < ind[num + 1] && nIdx >= ind[num + 1]) update(num, +1);
  if (oIdx > ind[num + 1] && nIdx <= ind[num + 1]) update(num, -1);
  if (oIdx < ind[num - 1] && nIdx >= ind[num - 1]) update(num - 1, -1);
  if (oIdx > ind[num - 1] && nIdx <= ind[num - 1]) update(num - 1, +1);
}
int main(void) {
  in = stdin;
  out = stdout;
  fscanf(in, "%d", &n);
  for (int i = 1; i <= n; i++) {
    fscanf(in, "%d", &a[i]);
    ind[a[i]] = i;
  }
  a[0] = 0;
  ind[0] = 0;
  a[n + 1] = n + 1;
  ind[n + 1] = n + 1;
  for (int i = 1; i <= n; i++) {
    if (ind[i + 1] < ind[i]) update(i, +1);
  }
  int numQueries;
  fscanf(in, "%d", &numQueries);
  for (int i = 0; i < numQueries; i++) {
    int t, x, y;
    fscanf(in, "%d %d %d", &t, &x, &y);
    if (t == 1) {
      fprintf(out, "%d\n", query(x, y - 1) + 1);
    } else {
      doUpdate(a[x], x, y);
      doUpdate(a[y], y, x);
      if (abs(a[x] - a[y]) == 1) {
        if (a[x] < a[y])
          update(a[x], -1);
        else
          update(a[y], +1);
      }
      swap(a[x], a[y]);
      swap(ind[a[x]], ind[a[y]]);
    }
  }
  return 0;
}

```