#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
const long long inf = 0x3f3f3f3f3f3f3f3f;
const double pi = acos(-1.0);
int a[300005];
int du[300005];
map<int, int> mp;
struct node {
  int l, r;
  int data;
  void _node(int ll, int rr, int dd) {
    l = ll;
    r = rr;
    data = dd;
  }
} tr[300005 * 4];
void build(int l, int r, int pos) {
  int mid = (l + r) / 2;
  tr[pos]._node(l, r, 0);
  if (l == r) return;
  build(l, mid, 2 * pos);
  build(mid + 1, r, 2 * pos + 1);
}
void update(int num, int pos, int del) {
  if (num >= tr[pos].l && num <= tr[pos].r) tr[pos].data += del;
  if (tr[pos].l == tr[pos].r) return;
  int l = tr[pos].l, r = tr[pos].r;
  int mid = (l + r) / 2;
  if (num <= mid)
    update(num, pos * 2, del);
  else
    update(num, pos * 2 + 1, del);
}
int read(int l, int r, int pos) {
  if (tr[pos].l == tr[pos].r || tr[pos].l == l && tr[pos].r == r)
    return tr[pos].data;
  int mid = (tr[pos].l + tr[pos].r) / 2;
  if (l > mid) {
    return read(l, r, pos * 2 + 1);
  } else if (r <= mid) {
    return read(l, r, pos * 2);
  } else {
    return read(l, mid, pos * 2) + read(mid + 1, r, pos * 2 + 1);
  }
}
int main() {
  int n;
  scanf("%d", &n);
  build(1, n, 1);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    mp[a[i]] = i;
  }
  update(1, 1, 1);
  for (int i = 2; i <= n; i++) {
    if (mp[i] >= mp[i - 1])
      du[i] = 1;
    else
      update(i, 1, 1);
  }
  int qnum;
  int pi = 0;
  scanf("%d", &qnum);
  for (int q = 1; q <= qnum; q++) {
    int o, l, r;
    scanf("%d%d%d", &o, &l, &r);
    if (l > r) swap(l, r);
    bool cf = false;
    if (r == l - 1) cf = true;
    if (o == 2) {
      int lsl = mp[a[l]], lsr = mp[a[r]];
      swap(a[l], a[r]);
      swap(mp[a[l]], mp[a[r]]);
      int mi = a[l], ma = a[r];
      if (mi > ma) swap(mi, ma);
      if (mi - 1 >= 1 && du[mi] == 0 && mp[mi - 1] < mp[mi]) {
        du[mi] = 1;
        update(mi, 1, -1);
      } else if (mi - 1 >= 1 && du[mi] && mp[mi] < mp[mi - 1]) {
        du[mi] = 0;
        update(mi, 1, 1);
      }
      if (mi + 1 <= n && du[mi + 1] == 0 && mp[mi] < mp[mi + 1]) {
        du[mi + 1] = 1;
        update(mi + 1, 1, -1);
      } else if (mi + 1 <= n && du[mi + 1] && mp[mi + 1] < mp[mi]) {
        du[mi + 1] = 0;
        update(mi + 1, 1, 1);
      }
      if (!cf && ma - 1 >= 1 && du[ma] == 0 && mp[ma - 1] < mp[ma]) {
        du[ma] = 1;
        update(ma, 1, -1);
      } else if (!cf && ma - 1 >= 1 && du[ma] && mp[ma] < mp[ma - 1]) {
        du[ma] = 0;
        update(ma, 1, 1);
      }
      if (ma + 1 <= n && du[ma + 1] == 0 && mp[ma] < mp[ma + 1]) {
        du[ma + 1] = 1;
        update(ma + 1, 1, -1);
      } else if (ma + 1 <= n && du[ma + 1] && mp[ma + 1] < mp[ma]) {
        du[ma + 1] = 0;
        update(ma + 1, 1, 1);
      }
    } else {
      pi++;
      int ans = read(l, r, 1);
      if (du[l] != 0) ans++;
      printf("%d\n", ans);
    }
  }
  return 0;
}

### Prompt

Generate 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>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
const long long inf = 0x3f3f3f3f3f3f3f3f;
const double pi = acos(-1.0);
int a[300005];
int du[300005];
map<int, int> mp;
struct node {
  int l, r;
  int data;
  void _node(int ll, int rr, int dd) {
    l = ll;
    r = rr;
    data = dd;
  }
} tr[300005 * 4];
void build(int l, int r, int pos) {
  int mid = (l + r) / 2;
  tr[pos]._node(l, r, 0);
  if (l == r) return;
  build(l, mid, 2 * pos);
  build(mid + 1, r, 2 * pos + 1);
}
void update(int num, int pos, int del) {
  if (num >= tr[pos].l && num <= tr[pos].r) tr[pos].data += del;
  if (tr[pos].l == tr[pos].r) return;
  int l = tr[pos].l, r = tr[pos].r;
  int mid = (l + r) / 2;
  if (num <= mid)
    update(num, pos * 2, del);
  else
    update(num, pos * 2 + 1, del);
}
int read(int l, int r, int pos) {
  if (tr[pos].l == tr[pos].r || tr[pos].l == l && tr[pos].r == r)
    return tr[pos].data;
  int mid = (tr[pos].l + tr[pos].r) / 2;
  if (l > mid) {
    return read(l, r, pos * 2 + 1);
  } else if (r <= mid) {
    return read(l, r, pos * 2);
  } else {
    return read(l, mid, pos * 2) + read(mid + 1, r, pos * 2 + 1);
  }
}
int main() {
  int n;
  scanf("%d", &n);
  build(1, n, 1);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    mp[a[i]] = i;
  }
  update(1, 1, 1);
  for (int i = 2; i <= n; i++) {
    if (mp[i] >= mp[i - 1])
      du[i] = 1;
    else
      update(i, 1, 1);
  }
  int qnum;
  int pi = 0;
  scanf("%d", &qnum);
  for (int q = 1; q <= qnum; q++) {
    int o, l, r;
    scanf("%d%d%d", &o, &l, &r);
    if (l > r) swap(l, r);
    bool cf = false;
    if (r == l - 1) cf = true;
    if (o == 2) {
      int lsl = mp[a[l]], lsr = mp[a[r]];
      swap(a[l], a[r]);
      swap(mp[a[l]], mp[a[r]]);
      int mi = a[l], ma = a[r];
      if (mi > ma) swap(mi, ma);
      if (mi - 1 >= 1 && du[mi] == 0 && mp[mi - 1] < mp[mi]) {
        du[mi] = 1;
        update(mi, 1, -1);
      } else if (mi - 1 >= 1 && du[mi] && mp[mi] < mp[mi - 1]) {
        du[mi] = 0;
        update(mi, 1, 1);
      }
      if (mi + 1 <= n && du[mi + 1] == 0 && mp[mi] < mp[mi + 1]) {
        du[mi + 1] = 1;
        update(mi + 1, 1, -1);
      } else if (mi + 1 <= n && du[mi + 1] && mp[mi + 1] < mp[mi]) {
        du[mi + 1] = 0;
        update(mi + 1, 1, 1);
      }
      if (!cf && ma - 1 >= 1 && du[ma] == 0 && mp[ma - 1] < mp[ma]) {
        du[ma] = 1;
        update(ma, 1, -1);
      } else if (!cf && ma - 1 >= 1 && du[ma] && mp[ma] < mp[ma - 1]) {
        du[ma] = 0;
        update(ma, 1, 1);
      }
      if (ma + 1 <= n && du[ma + 1] == 0 && mp[ma] < mp[ma + 1]) {
        du[ma + 1] = 1;
        update(ma + 1, 1, -1);
      } else if (ma + 1 <= n && du[ma + 1] && mp[ma + 1] < mp[ma]) {
        du[ma + 1] = 0;
        update(ma + 1, 1, 1);
      }
    } else {
      pi++;
      int ans = read(l, r, 1);
      if (du[l] != 0) ans++;
      printf("%d\n", ans);
    }
  }
  return 0;
}

```