#include <bits/stdc++.h>
using namespace std;
int n, m, a[300005], Rank[300005];
int c[300005];
void add(int pos, int val) {
  if (pos == 0) return;
  while (pos <= n) {
    c[pos] += val;
    pos += (pos & -pos);
  }
}
int getsum(int pos) {
  int ans = 0;
  while (pos) {
    ans += c[pos];
    pos -= (pos & -pos);
  }
  return ans;
}
int calc(int l, int r) { return getsum(r - 1) - getsum(l - 1) + 1; }
void debug(int n) {
  int i;
  printf("******************debug*****************\n");
  for (i = 1; i <= n; i++) printf("%d ", a[i]);
  printf("\n");
  for (i = 1; i <= n; i++) printf("%d ", Rank[i]);
  printf("\n");
  printf("******************debug*****************\n");
}
int main() {
  int i, op, x, y, vx, vy, rx, ry, cc;
  while (scanf("%d", &n) != EOF) {
    memset(c, 0, sizeof(c));
    for (i = 1; i <= n; i++) {
      scanf("%d", &a[i]);
      Rank[a[i]] = i;
    }
    for (i = 1; i < n; i++)
      if (Rank[i] > Rank[i + 1]) add(i, 1);
    scanf("%d", &m);
    for (i = 1; i <= m; i++) {
      scanf("%d %d %d", &op, &x, &y);
      if (op == 1)
        printf("%d\n", calc(x, y));
      else {
        vx = a[x], vy = a[y];
        if (vx > vy) swap(vx, vy);
        rx = Rank[vx], ry = Rank[vy];
        cc = (ry < Rank[vx - 1]) - (rx < Rank[vx - 1]);
        add(vx - 1, cc);
        cc = (rx > Rank[vy + 1]) - (ry > Rank[vy + 1]);
        add(vy, cc);
        if (abs(vx - vy) == 1) {
          if (rx > ry)
            add(vx, -1);
          else
            add(vx, 1);
        } else {
          cc = (ry > Rank[vx + 1]) - (rx > Rank[vx + 1]);
          add(vx, cc);
          cc = (rx < Rank[vy - 1]) - (ry < Rank[vy - 1]);
          add(vy - 1, cc);
        }
        swap(a[x], a[y]);
        swap(Rank[vx], Rank[vy]);
      }
    }
  }
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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;
int n, m, a[300005], Rank[300005];
int c[300005];
void add(int pos, int val) {
  if (pos == 0) return;
  while (pos <= n) {
    c[pos] += val;
    pos += (pos & -pos);
  }
}
int getsum(int pos) {
  int ans = 0;
  while (pos) {
    ans += c[pos];
    pos -= (pos & -pos);
  }
  return ans;
}
int calc(int l, int r) { return getsum(r - 1) - getsum(l - 1) + 1; }
void debug(int n) {
  int i;
  printf("******************debug*****************\n");
  for (i = 1; i <= n; i++) printf("%d ", a[i]);
  printf("\n");
  for (i = 1; i <= n; i++) printf("%d ", Rank[i]);
  printf("\n");
  printf("******************debug*****************\n");
}
int main() {
  int i, op, x, y, vx, vy, rx, ry, cc;
  while (scanf("%d", &n) != EOF) {
    memset(c, 0, sizeof(c));
    for (i = 1; i <= n; i++) {
      scanf("%d", &a[i]);
      Rank[a[i]] = i;
    }
    for (i = 1; i < n; i++)
      if (Rank[i] > Rank[i + 1]) add(i, 1);
    scanf("%d", &m);
    for (i = 1; i <= m; i++) {
      scanf("%d %d %d", &op, &x, &y);
      if (op == 1)
        printf("%d\n", calc(x, y));
      else {
        vx = a[x], vy = a[y];
        if (vx > vy) swap(vx, vy);
        rx = Rank[vx], ry = Rank[vy];
        cc = (ry < Rank[vx - 1]) - (rx < Rank[vx - 1]);
        add(vx - 1, cc);
        cc = (rx > Rank[vy + 1]) - (ry > Rank[vy + 1]);
        add(vy, cc);
        if (abs(vx - vy) == 1) {
          if (rx > ry)
            add(vx, -1);
          else
            add(vx, 1);
        } else {
          cc = (ry > Rank[vx + 1]) - (rx > Rank[vx + 1]);
          add(vx, cc);
          cc = (rx < Rank[vy - 1]) - (ry < Rank[vy - 1]);
          add(vy - 1, cc);
        }
        swap(a[x], a[y]);
        swap(Rank[vx], Rank[vy]);
      }
    }
  }
  return 0;
}

```