#include <bits/stdc++.h>
using namespace std;
const int maxn = 300005;
int a[maxn], ps[maxn], sum[maxn << 2];
void build(int l, int r, int rt) {
  if (l == r) {
    sum[rt] = ps[l + 1] < ps[l];
    return;
  }
  int m = (l + r) >> 1;
  build(l, m, rt << 1);
  build(m + 1, r, rt << 1 | 1);
  sum[rt] = sum[rt << 1] + sum[rt << 1 | 1];
}
void update(int p, int l, int r, int rt) {
  if (l == r) {
    sum[rt] = ps[l + 1] < ps[l];
    return;
  }
  int m = (l + r) >> 1;
  if (p <= m)
    update(p, l, m, rt << 1);
  else
    update(p, m + 1, r, rt << 1 | 1);
  sum[rt] = sum[rt << 1] + sum[rt << 1 | 1];
}
int query(int L, int R, int l, int r, int rt) {
  if (L <= l && r <= R) return sum[rt];
  int m = (l + r) >> 1, ret = 0;
  if (L <= m) ret += query(L, R, l, m, rt << 1);
  if (m < R) ret += query(L, R, m + 1, r, rt << 1 | 1);
  return ret;
}
int main() {
  int i, j, m, n, l, r, x, y;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) scanf("%d", &a[i]), ps[a[i]] = i;
  build(1, n - 1, 1);
  scanf("%d", &m);
  while (m--) {
    scanf("%d%d%d", &x, &l, &r);
    if (x == 1) {
      if (l != r)
        printf("%d\n", 1 + query(l, r - 1, 1, n - 1, 1));
      else
        printf("1\n");
    } else {
      if (l == r) continue;
      swap(ps[a[l]], ps[a[r]]);
      swap(a[l], a[r]);
      x = min(a[l], a[r]), y = max(a[l], a[r]);
      if (x > 1) update(x - 1, 1, n - 1, 1);
      update(x, 1, n - 1, 1);
      update(y - 1, 1, n - 1, 1);
      if (y < n) update(y, 1, n - 1, 1);
    }
  }
  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 maxn = 300005;
int a[maxn], ps[maxn], sum[maxn << 2];
void build(int l, int r, int rt) {
  if (l == r) {
    sum[rt] = ps[l + 1] < ps[l];
    return;
  }
  int m = (l + r) >> 1;
  build(l, m, rt << 1);
  build(m + 1, r, rt << 1 | 1);
  sum[rt] = sum[rt << 1] + sum[rt << 1 | 1];
}
void update(int p, int l, int r, int rt) {
  if (l == r) {
    sum[rt] = ps[l + 1] < ps[l];
    return;
  }
  int m = (l + r) >> 1;
  if (p <= m)
    update(p, l, m, rt << 1);
  else
    update(p, m + 1, r, rt << 1 | 1);
  sum[rt] = sum[rt << 1] + sum[rt << 1 | 1];
}
int query(int L, int R, int l, int r, int rt) {
  if (L <= l && r <= R) return sum[rt];
  int m = (l + r) >> 1, ret = 0;
  if (L <= m) ret += query(L, R, l, m, rt << 1);
  if (m < R) ret += query(L, R, m + 1, r, rt << 1 | 1);
  return ret;
}
int main() {
  int i, j, m, n, l, r, x, y;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) scanf("%d", &a[i]), ps[a[i]] = i;
  build(1, n - 1, 1);
  scanf("%d", &m);
  while (m--) {
    scanf("%d%d%d", &x, &l, &r);
    if (x == 1) {
      if (l != r)
        printf("%d\n", 1 + query(l, r - 1, 1, n - 1, 1));
      else
        printf("1\n");
    } else {
      if (l == r) continue;
      swap(ps[a[l]], ps[a[r]]);
      swap(a[l], a[r]);
      x = min(a[l], a[r]), y = max(a[l], a[r]);
      if (x > 1) update(x - 1, 1, n - 1, 1);
      update(x, 1, n - 1, 1);
      update(y - 1, 1, n - 1, 1);
      if (y < n) update(y, 1, n - 1, 1);
    }
  }
  return 0;
}

```