#include <bits/stdc++.h>
using namespace std;
const int N = 4 * 100000;
int a[N], p[N];
int n;
struct node {
  int l, r;
  int w;
} t[N * 4];
void init(int l, int r, int rt) {
  t[rt].l = l;
  t[rt].r = r;
  int mid = (l + r) / 2;
  if (l == r) {
    if (p[l] < p[l + 1])
      t[rt].w = 0;
    else
      t[rt].w = 1;
    return;
  }
  init(l, mid, rt * 2);
  init(mid + 1, r, rt * 2 + 1);
  t[rt].w = t[rt * 2].w + t[rt * 2 + 1].w;
}
int query(int l, int r, int rt) {
  if (t[rt].l == l && t[rt].r == r) {
    return t[rt].w;
  }
  int mid = (t[rt].r + t[rt].l) / 2;
  if (r <= mid) {
    return query(l, r, rt * 2);
  } else if (l > mid) {
    return query(l, r, rt * 2 + 1);
  } else {
    return query(l, mid, rt * 2) + query(mid + 1, r, rt * 2 + 1);
  }
}
void update(int l, int v, int rt) {
  if (t[rt].r == t[rt].l) {
    t[rt].w += v;
    return;
  }
  t[rt].w += v;
  int mid = (t[rt].l + t[rt].r) / 2;
  if (l <= mid)
    update(l, v, rt * 2);
  else
    update(l, v, rt * 2 + 1);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    p[a[i]] = i;
  }
  p[0] = 0;
  p[n + 1] = n;
  init(1, n, 1);
  int Q;
  scanf("%d", &Q);
  while (Q--) {
    int type, l, r;
    scanf("%d%d%d", &type, &l, &r);
    if (type == 1) {
      printf("%d\n", query(l, r - 1, 1) + 1);
    } else {
      if (a[l] > a[r]) swap(l, r);
      int x[4], y[4];
      x[0] = a[l] - 1;
      x[1] = a[l];
      x[2] = a[r] - 1;
      x[3] = a[r];
      for (int i = 0; i < 4; i++) {
        if (p[x[i]] < p[x[i] + 1])
          y[i] = 0;
        else
          y[i] = 1;
      }
      swap(p[a[l]], p[a[r]]);
      swap(a[l], a[r]);
      for (int i = 0; i < 4; i++) {
        if (p[x[i]] < p[x[i] + 1])
          y[i] = -y[i];
        else
          y[i] = 1 - y[i];
        if (i == 2 && x[i] == x[i - 1]) continue;
        if (y[i] == 0) continue;
        update(x[i], y[i], 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 N = 4 * 100000;
int a[N], p[N];
int n;
struct node {
  int l, r;
  int w;
} t[N * 4];
void init(int l, int r, int rt) {
  t[rt].l = l;
  t[rt].r = r;
  int mid = (l + r) / 2;
  if (l == r) {
    if (p[l] < p[l + 1])
      t[rt].w = 0;
    else
      t[rt].w = 1;
    return;
  }
  init(l, mid, rt * 2);
  init(mid + 1, r, rt * 2 + 1);
  t[rt].w = t[rt * 2].w + t[rt * 2 + 1].w;
}
int query(int l, int r, int rt) {
  if (t[rt].l == l && t[rt].r == r) {
    return t[rt].w;
  }
  int mid = (t[rt].r + t[rt].l) / 2;
  if (r <= mid) {
    return query(l, r, rt * 2);
  } else if (l > mid) {
    return query(l, r, rt * 2 + 1);
  } else {
    return query(l, mid, rt * 2) + query(mid + 1, r, rt * 2 + 1);
  }
}
void update(int l, int v, int rt) {
  if (t[rt].r == t[rt].l) {
    t[rt].w += v;
    return;
  }
  t[rt].w += v;
  int mid = (t[rt].l + t[rt].r) / 2;
  if (l <= mid)
    update(l, v, rt * 2);
  else
    update(l, v, rt * 2 + 1);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    p[a[i]] = i;
  }
  p[0] = 0;
  p[n + 1] = n;
  init(1, n, 1);
  int Q;
  scanf("%d", &Q);
  while (Q--) {
    int type, l, r;
    scanf("%d%d%d", &type, &l, &r);
    if (type == 1) {
      printf("%d\n", query(l, r - 1, 1) + 1);
    } else {
      if (a[l] > a[r]) swap(l, r);
      int x[4], y[4];
      x[0] = a[l] - 1;
      x[1] = a[l];
      x[2] = a[r] - 1;
      x[3] = a[r];
      for (int i = 0; i < 4; i++) {
        if (p[x[i]] < p[x[i] + 1])
          y[i] = 0;
        else
          y[i] = 1;
      }
      swap(p[a[l]], p[a[r]]);
      swap(a[l], a[r]);
      for (int i = 0; i < 4; i++) {
        if (p[x[i]] < p[x[i] + 1])
          y[i] = -y[i];
        else
          y[i] = 1 - y[i];
        if (i == 2 && x[i] == x[i - 1]) continue;
        if (y[i] == 0) continue;
        update(x[i], y[i], 1);
      }
    }
  }
  return 0;
}

```