#include <bits/stdc++.h>
int a[3 * 100010], pos[3 * 100010];
int idx[3 * 100010];
int n;
int lowbit(int x) { return x & (-x); }
int Sum(int x) {
  int sum = 0;
  while (x) sum += idx[x], x -= lowbit(x);
  return sum;
}
void add(int pos, int num) {
  while (pos <= n) idx[pos] += num, pos += lowbit(pos);
}
void init() {
  int i;
  memset(idx, 0, sizeof(idx));
  for (i = 1; i <= n; i++)
    if (pos[i] < pos[i - 1]) add(i, 1);
}
void update(int p, int x) {
  if (pos[p] < pos[p - 1]) add(p, -1);
  if (x < pos[p - 1]) add(p, 1);
  if (p == n) return;
  if (pos[p + 1] < pos[p]) add(p + 1, -1);
  if (pos[p + 1] < x) add(p + 1, 1);
}
void change(int x, int y) {
  int xx, yy, px, py;
  xx = a[x], yy = a[y];
  a[x] = yy, a[y] = xx;
  update(xx, y);
  pos[xx] = y;
  update(yy, x);
  pos[yy] = x;
}
int main() {
  int i, q, t, x, y;
  while (~scanf("%d", &n)) {
    for (i = 1; i <= n; i++) scanf("%d", &a[i]), pos[a[i]] = i;
    init();
    scanf("%d", &q);
    while (q--) {
      scanf("%d%d%d", &t, &x, &y);
      if (t == 1)
        printf("%d\n", Sum(y) - Sum(x) + 1);
      else
        change(x, y);
    }
  }
  return 0;
}

### Prompt

Please provide a CPP coded solution 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>
int a[3 * 100010], pos[3 * 100010];
int idx[3 * 100010];
int n;
int lowbit(int x) { return x & (-x); }
int Sum(int x) {
  int sum = 0;
  while (x) sum += idx[x], x -= lowbit(x);
  return sum;
}
void add(int pos, int num) {
  while (pos <= n) idx[pos] += num, pos += lowbit(pos);
}
void init() {
  int i;
  memset(idx, 0, sizeof(idx));
  for (i = 1; i <= n; i++)
    if (pos[i] < pos[i - 1]) add(i, 1);
}
void update(int p, int x) {
  if (pos[p] < pos[p - 1]) add(p, -1);
  if (x < pos[p - 1]) add(p, 1);
  if (p == n) return;
  if (pos[p + 1] < pos[p]) add(p + 1, -1);
  if (pos[p + 1] < x) add(p + 1, 1);
}
void change(int x, int y) {
  int xx, yy, px, py;
  xx = a[x], yy = a[y];
  a[x] = yy, a[y] = xx;
  update(xx, y);
  pos[xx] = y;
  update(yy, x);
  pos[yy] = x;
}
int main() {
  int i, q, t, x, y;
  while (~scanf("%d", &n)) {
    for (i = 1; i <= n; i++) scanf("%d", &a[i]), pos[a[i]] = i;
    init();
    scanf("%d", &q);
    while (q--) {
      scanf("%d%d%d", &t, &x, &y);
      if (t == 1)
        printf("%d\n", Sum(y) - Sum(x) + 1);
      else
        change(x, y);
    }
  }
  return 0;
}

```