#include <bits/stdc++.h>
using namespace std;
const int loveisblue = 486;
const int maxn = 300086;
const int maxm = 100086;
const int inf = 0x3f3f3f3f;
const long long Inf = 999999999999999999;
const int mod = 1000000007;
const double eps = 1e-6;
const double pi = acos(-1);
int num[maxn], pos[maxn];
int bit[maxn];
int lowbit(int x) { return x & -x; }
int query(int pos) {
  int ans = 0;
  while (pos) {
    ans += bit[pos];
    pos -= lowbit(pos);
  }
  return ans;
}
void update(int pos, int val) {
  while (pos < maxn) {
    bit[pos] += val;
    pos += lowbit(pos);
  }
}
int main() {
  ios::sync_with_stdio(true);
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &num[i]);
    pos[num[i]] = i;
  }
  for (int i = 1; i < n; i++) {
    if (pos[i] > pos[i + 1]) {
      update(i, 1);
    }
  }
  int m;
  scanf("%d", &m);
  while (m--) {
    int type;
    scanf("%d", &type);
    if (type == 1) {
      int l, r;
      scanf("%d%d", &l, &r);
      printf("%d\n", query(r - 1) - query(l - 1) + 1);
    } else {
      int x, y;
      scanf("%d%d", &x, &y);
      swap(num[x], num[y]);
      x = num[x];
      y = num[y];
      int t1 = pos[x];
      int t2 = pos[y];
      if (t1 > t2) {
        swap(x, y);
        swap(t1, t2);
      }
      if (pos[x + 1] >= pos[x] && pos[x + 1] <= pos[y]) {
        update(x, 1);
      }
      if (x > 1 && x - 1 != y && pos[x - 1] >= pos[x] && pos[x - 1] <= pos[y]) {
        update(x - 1, -1);
      }
      if (pos[y + 1] >= pos[x] && pos[y + 1] <= pos[y]) {
        update(y, -1);
      }
      if (y > 1 && y - 1 != x && pos[y - 1] >= pos[x] && pos[y - 1] <= pos[y]) {
        update(y - 1, 1);
      }
      swap(pos[x], pos[y]);
    }
  }
  return 0;
}

### Prompt

Your challenge is to write 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 loveisblue = 486;
const int maxn = 300086;
const int maxm = 100086;
const int inf = 0x3f3f3f3f;
const long long Inf = 999999999999999999;
const int mod = 1000000007;
const double eps = 1e-6;
const double pi = acos(-1);
int num[maxn], pos[maxn];
int bit[maxn];
int lowbit(int x) { return x & -x; }
int query(int pos) {
  int ans = 0;
  while (pos) {
    ans += bit[pos];
    pos -= lowbit(pos);
  }
  return ans;
}
void update(int pos, int val) {
  while (pos < maxn) {
    bit[pos] += val;
    pos += lowbit(pos);
  }
}
int main() {
  ios::sync_with_stdio(true);
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &num[i]);
    pos[num[i]] = i;
  }
  for (int i = 1; i < n; i++) {
    if (pos[i] > pos[i + 1]) {
      update(i, 1);
    }
  }
  int m;
  scanf("%d", &m);
  while (m--) {
    int type;
    scanf("%d", &type);
    if (type == 1) {
      int l, r;
      scanf("%d%d", &l, &r);
      printf("%d\n", query(r - 1) - query(l - 1) + 1);
    } else {
      int x, y;
      scanf("%d%d", &x, &y);
      swap(num[x], num[y]);
      x = num[x];
      y = num[y];
      int t1 = pos[x];
      int t2 = pos[y];
      if (t1 > t2) {
        swap(x, y);
        swap(t1, t2);
      }
      if (pos[x + 1] >= pos[x] && pos[x + 1] <= pos[y]) {
        update(x, 1);
      }
      if (x > 1 && x - 1 != y && pos[x - 1] >= pos[x] && pos[x - 1] <= pos[y]) {
        update(x - 1, -1);
      }
      if (pos[y + 1] >= pos[x] && pos[y + 1] <= pos[y]) {
        update(y, -1);
      }
      if (y > 1 && y - 1 != x && pos[y - 1] >= pos[x] && pos[y - 1] <= pos[y]) {
        update(y - 1, 1);
      }
      swap(pos[x], pos[y]);
    }
  }
  return 0;
}

```