#include <bits/stdc++.h>
using namespace std;
struct Node {
  int L, R, sum;
} node[100100 << 4];
int n, m;
void build_tree(int l, int r, int id) {
  node[id].L = l;
  node[id].R = r;
  node[id].sum = 0;
  int mid = (l + r) / 2;
  if (l == r) return;
  build_tree(l, mid, id << 1);
  build_tree(mid + 1, r, (id << 1) + 1);
}
void update(int id, int pt, int v) {
  if (node[id].L == node[id].R) {
    node[id].sum = v;
  } else {
    int mid = (node[id].L + node[id].R) / 2;
    if (mid >= pt)
      update(id << 1, pt, v);
    else
      update((id << 1) + 1, pt, v);
    node[id].sum = node[(id << 1) + 1].sum + node[id << 1].sum;
  }
}
int query(int l, int r, int id) {
  if (node[id].L == l && node[id].R == r)
    return node[id].sum;
  else {
    int mid = (node[id].L + node[id].R) / 2;
    if (mid >= r)
      return query(l, r, id << 1);
    else if (mid < l)
      return query(l, r, (id << 1) + 1);
    else
      return query(l, mid, id << 1) + query(mid + 1, r, (id << 1) + 1);
  }
}
int p[100100 * 4], P[100100 * 4];
int ans;
int q, x, y;
int main() {
  while (~scanf("%d", &n)) {
    build_tree(1, n, 1);
    for (int i = 1; i <= n; i++) {
      scanf("%d", &p[i]);
      P[p[i]] = i;
    }
    for (int i = 1; i < n; i++) {
      if (P[i] > P[i + 1]) {
        update(1, i + 1, 1);
      }
    }
    scanf("%d", &m);
    for (int i = 1; i <= m; i++) {
      scanf("%d%d%d", &q, &x, &y);
      if (q == 1) {
        ans = query(x + 1, y, 1);
        printf("%d\n", ans + 1);
      } else {
        int a, b;
        a = p[x];
        b = p[y];
        p[x] = b;
        p[y] = a;
        P[a] = y;
        P[b] = x;
        int tmp;
        if (a != n) {
          tmp = a + 1;
          if (P[tmp] < P[a]) update(1, tmp, 1);
        }
        if (a != 1) {
          tmp = a - 1;
          if (P[tmp] < P[a]) update(1, a, 0);
        }
        if (b != n) {
          tmp = b + 1;
          if (P[b] < P[tmp]) update(1, tmp, 0);
        }
        if (b != 1) {
          tmp = b - 1;
          if (P[b] < P[tmp]) update(1, b, 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;
struct Node {
  int L, R, sum;
} node[100100 << 4];
int n, m;
void build_tree(int l, int r, int id) {
  node[id].L = l;
  node[id].R = r;
  node[id].sum = 0;
  int mid = (l + r) / 2;
  if (l == r) return;
  build_tree(l, mid, id << 1);
  build_tree(mid + 1, r, (id << 1) + 1);
}
void update(int id, int pt, int v) {
  if (node[id].L == node[id].R) {
    node[id].sum = v;
  } else {
    int mid = (node[id].L + node[id].R) / 2;
    if (mid >= pt)
      update(id << 1, pt, v);
    else
      update((id << 1) + 1, pt, v);
    node[id].sum = node[(id << 1) + 1].sum + node[id << 1].sum;
  }
}
int query(int l, int r, int id) {
  if (node[id].L == l && node[id].R == r)
    return node[id].sum;
  else {
    int mid = (node[id].L + node[id].R) / 2;
    if (mid >= r)
      return query(l, r, id << 1);
    else if (mid < l)
      return query(l, r, (id << 1) + 1);
    else
      return query(l, mid, id << 1) + query(mid + 1, r, (id << 1) + 1);
  }
}
int p[100100 * 4], P[100100 * 4];
int ans;
int q, x, y;
int main() {
  while (~scanf("%d", &n)) {
    build_tree(1, n, 1);
    for (int i = 1; i <= n; i++) {
      scanf("%d", &p[i]);
      P[p[i]] = i;
    }
    for (int i = 1; i < n; i++) {
      if (P[i] > P[i + 1]) {
        update(1, i + 1, 1);
      }
    }
    scanf("%d", &m);
    for (int i = 1; i <= m; i++) {
      scanf("%d%d%d", &q, &x, &y);
      if (q == 1) {
        ans = query(x + 1, y, 1);
        printf("%d\n", ans + 1);
      } else {
        int a, b;
        a = p[x];
        b = p[y];
        p[x] = b;
        p[y] = a;
        P[a] = y;
        P[b] = x;
        int tmp;
        if (a != n) {
          tmp = a + 1;
          if (P[tmp] < P[a]) update(1, tmp, 1);
        }
        if (a != 1) {
          tmp = a - 1;
          if (P[tmp] < P[a]) update(1, a, 0);
        }
        if (b != n) {
          tmp = b + 1;
          if (P[b] < P[tmp]) update(1, tmp, 0);
        }
        if (b != 1) {
          tmp = b - 1;
          if (P[b] < P[tmp]) update(1, b, 1);
        }
      }
    }
  }
  return 0;
}

```