#include <bits/stdc++.h>
using namespace std;
struct tree {
  int sum;
  int pre, suff;
};
tree m[4 * 300100];
int n, q;
int a[300100], pos[300100];
void init(int node, int b, int e);
void update(int node, int b, int e, int k);
int query(int node, int b, int e, int x, int y);
int main() {
  int i, j, k, l, v, test, t = 1;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    pos[a[i]] = i;
  }
  init(1, 1, n);
  scanf("%d", &q);
  int type, r1, r2;
  for (int i1 = 1; i1 <= q; i1++) {
    scanf("%d %d %d", &type, &r1, &r2);
    if (type == 1) {
      int ans = 1, last = pos[r1];
      ans = query(1, 1, n, r1, r2);
      printf("%d\n", ans);
    } else {
      swap(a[r1], a[r2]);
      swap(pos[a[r1]], pos[a[r2]]);
      update(1, 1, n, a[r1]);
      update(1, 1, n, a[r2]);
      update(1, 1, n, a[r1]);
    }
  }
  return 0;
}
int query(int node, int b, int e, int x, int y) {
  if (b == e) {
    return m[node].sum;
  }
  int k = e - b + 1, l;
  if (b == x && e == y) {
    return m[node].sum;
  }
  int left, right, mid, ret = 0;
  left = node << 1;
  right = left + 1;
  mid = b + e;
  mid /= 2;
  if (y <= mid) {
    return query(left, b, mid, x, y);
  } else if (x > mid) {
    return query(right, mid + 1, e, x, y);
  } else {
    ret += query(left, b, mid, x, mid);
    ret += query(right, mid + 1, e, mid + 1, y);
  }
  if (pos[mid] < pos[mid + 1]) ret--;
  return ret;
}
void update(int node, int b, int e, int k) {
  if (b == e) {
    m[node].sum = 1;
    m[node].pre = pos[b];
    m[node].suff = pos[e];
    return;
  }
  int left, right, mid;
  left = node << 1;
  right = left + 1;
  mid = b + e;
  mid /= 2;
  if (k <= mid)
    update(left, b, mid, k);
  else
    update(right, mid + 1, e, k);
  m[node].pre = pos[b];
  m[node].suff = pos[e];
  m[node].sum = m[left].sum + m[right].sum;
  if (m[left].suff < m[right].pre) {
    m[node].sum--;
  }
}
void init(int node, int b, int e) {
  if (b == e) {
    m[node].sum = 1;
    m[node].pre = pos[b];
    m[node].suff = pos[e];
    return;
  }
  int left, right, mid;
  left = node << 1;
  right = left + 1;
  mid = b + e;
  mid /= 2;
  init(left, b, mid);
  init(right, mid + 1, e);
  m[node].pre = pos[b];
  m[node].suff = pos[e];
  m[node].sum = m[left].sum + m[right].sum;
  if (m[left].suff < m[right].pre) {
    m[node].sum--;
  }
}

### Prompt

Please formulate 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 tree {
  int sum;
  int pre, suff;
};
tree m[4 * 300100];
int n, q;
int a[300100], pos[300100];
void init(int node, int b, int e);
void update(int node, int b, int e, int k);
int query(int node, int b, int e, int x, int y);
int main() {
  int i, j, k, l, v, test, t = 1;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    pos[a[i]] = i;
  }
  init(1, 1, n);
  scanf("%d", &q);
  int type, r1, r2;
  for (int i1 = 1; i1 <= q; i1++) {
    scanf("%d %d %d", &type, &r1, &r2);
    if (type == 1) {
      int ans = 1, last = pos[r1];
      ans = query(1, 1, n, r1, r2);
      printf("%d\n", ans);
    } else {
      swap(a[r1], a[r2]);
      swap(pos[a[r1]], pos[a[r2]]);
      update(1, 1, n, a[r1]);
      update(1, 1, n, a[r2]);
      update(1, 1, n, a[r1]);
    }
  }
  return 0;
}
int query(int node, int b, int e, int x, int y) {
  if (b == e) {
    return m[node].sum;
  }
  int k = e - b + 1, l;
  if (b == x && e == y) {
    return m[node].sum;
  }
  int left, right, mid, ret = 0;
  left = node << 1;
  right = left + 1;
  mid = b + e;
  mid /= 2;
  if (y <= mid) {
    return query(left, b, mid, x, y);
  } else if (x > mid) {
    return query(right, mid + 1, e, x, y);
  } else {
    ret += query(left, b, mid, x, mid);
    ret += query(right, mid + 1, e, mid + 1, y);
  }
  if (pos[mid] < pos[mid + 1]) ret--;
  return ret;
}
void update(int node, int b, int e, int k) {
  if (b == e) {
    m[node].sum = 1;
    m[node].pre = pos[b];
    m[node].suff = pos[e];
    return;
  }
  int left, right, mid;
  left = node << 1;
  right = left + 1;
  mid = b + e;
  mid /= 2;
  if (k <= mid)
    update(left, b, mid, k);
  else
    update(right, mid + 1, e, k);
  m[node].pre = pos[b];
  m[node].suff = pos[e];
  m[node].sum = m[left].sum + m[right].sum;
  if (m[left].suff < m[right].pre) {
    m[node].sum--;
  }
}
void init(int node, int b, int e) {
  if (b == e) {
    m[node].sum = 1;
    m[node].pre = pos[b];
    m[node].suff = pos[e];
    return;
  }
  int left, right, mid;
  left = node << 1;
  right = left + 1;
  mid = b + e;
  mid /= 2;
  init(left, b, mid);
  init(right, mid + 1, e);
  m[node].pre = pos[b];
  m[node].suff = pos[e];
  m[node].sum = m[left].sum + m[right].sum;
  if (m[left].suff < m[right].pre) {
    m[node].sum--;
  }
}

```