#include <bits/stdc++.h>
using namespace std;
const int MAXN = 300010;
int a[MAXN];
int b[MAXN];
int n;
int dp[110][110];
int solve(int x, int y) {
  if (x == y) return 0;
  int t1 = b[x];
  int t2 = b[y];
  int p1 = x;
  int p2 = y;
  while (b[p1 + 1] > b[p1] && p1 + 1 <= p2) p1++;
  while (b[p2 - 1] < b[p2] && p1 + 1 <= p2) p2--;
  if (p1 == p2) return 0;
  return 1 + solve(p1 + 1, p2);
  int ans1, ans2;
  if (b[x + 1] > t1)
    ans1 = solve(x + 1, y);
  else
    ans1 = 1 + solve(x + 1, y);
  if (b[y - 1] < t2)
    ans2 = solve(x, y - 1);
  else
    ans2 = 1 + solve(x, y - 1);
  return min(ans1, ans2);
}
int main() {
  int m;
  int p, x, y;
  while (scanf("%d", &n) == 1) {
    for (int i = 1; i <= n; i++) {
      scanf("%d", &a[i]);
      b[a[i]] = i;
    }
    scanf("%d", &m);
    while (m--) {
      scanf("%d%d%d", &p, &x, &y);
      if (p == 2) {
        swap(a[x], a[y]);
        b[a[x]] = x;
        b[a[y]] = y;
      } else {
        printf("%d\n", solve(x, y) + 1);
      }
    }
  }
  return 0;
}

### Prompt

Please provide a CPP coded solution 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;
const int MAXN = 300010;
int a[MAXN];
int b[MAXN];
int n;
int dp[110][110];
int solve(int x, int y) {
  if (x == y) return 0;
  int t1 = b[x];
  int t2 = b[y];
  int p1 = x;
  int p2 = y;
  while (b[p1 + 1] > b[p1] && p1 + 1 <= p2) p1++;
  while (b[p2 - 1] < b[p2] && p1 + 1 <= p2) p2--;
  if (p1 == p2) return 0;
  return 1 + solve(p1 + 1, p2);
  int ans1, ans2;
  if (b[x + 1] > t1)
    ans1 = solve(x + 1, y);
  else
    ans1 = 1 + solve(x + 1, y);
  if (b[y - 1] < t2)
    ans2 = solve(x, y - 1);
  else
    ans2 = 1 + solve(x, y - 1);
  return min(ans1, ans2);
}
int main() {
  int m;
  int p, x, y;
  while (scanf("%d", &n) == 1) {
    for (int i = 1; i <= n; i++) {
      scanf("%d", &a[i]);
      b[a[i]] = i;
    }
    scanf("%d", &m);
    while (m--) {
      scanf("%d%d%d", &p, &x, &y);
      if (p == 2) {
        swap(a[x], a[y]);
        b[a[x]] = x;
        b[a[y]] = y;
      } else {
        printf("%d\n", solve(x, y) + 1);
      }
    }
  }
  return 0;
}

```