#include <bits/stdc++.h>
using namespace std;
template <class T>
bool checkMin(T& a, T b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
bool checkMax(T& a, T b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
const int INF = 1e9;
const int MAXN = 3e5 + 10;
int n, q;
int A[MAXN];
int p[MAXN];
int ft[MAXN + 1];
void update(int i, int x) {
  if (i != 0)
    while (i <= MAXN) {
      ft[i] += x;
      i += i & -i;
    }
}
int query(int t) {
  int sum = 0;
  while (t > 0) {
    sum += ft[t];
    t -= t & -t;
  }
  return sum;
}
int query(int i, int j) { return query(j) - query(i - 1); }
int main() {
  while (scanf("%d", &n) != EOF) {
    memset(ft, 0, sizeof ft);
    for (int i = (1); i <= (n); i++) {
      scanf("%d", A + i);
      p[A[i]] = i;
    }
    for (int i = (2); i <= (n); i++)
      if (p[i - 1] > p[i]) update(i - 1, 1);
    scanf("%d", &q);
    while (q--) {
      int qtype, u, v;
      scanf("%d%d%d", &qtype, &u, &v);
      if (qtype == 2) {
        p[A[u]] = v;
        p[A[v]] = u;
        swap(A[u], A[v]);
        for (int i = 0; i < 2; ++i) {
          for (int d = 0; d <= 1; d++) {
            int a = A[i ? v : u] + d;
            if (a > 1 && a <= n) {
              update(a - 1, (p[a - 1] > p[a]) - query(a - 1, a - 1));
            }
          }
        }
      } else {
        printf("%d\n", query(u, v - 1) + 1);
      }
    }
  }
}

### 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;
template <class T>
bool checkMin(T& a, T b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
bool checkMax(T& a, T b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
const int INF = 1e9;
const int MAXN = 3e5 + 10;
int n, q;
int A[MAXN];
int p[MAXN];
int ft[MAXN + 1];
void update(int i, int x) {
  if (i != 0)
    while (i <= MAXN) {
      ft[i] += x;
      i += i & -i;
    }
}
int query(int t) {
  int sum = 0;
  while (t > 0) {
    sum += ft[t];
    t -= t & -t;
  }
  return sum;
}
int query(int i, int j) { return query(j) - query(i - 1); }
int main() {
  while (scanf("%d", &n) != EOF) {
    memset(ft, 0, sizeof ft);
    for (int i = (1); i <= (n); i++) {
      scanf("%d", A + i);
      p[A[i]] = i;
    }
    for (int i = (2); i <= (n); i++)
      if (p[i - 1] > p[i]) update(i - 1, 1);
    scanf("%d", &q);
    while (q--) {
      int qtype, u, v;
      scanf("%d%d%d", &qtype, &u, &v);
      if (qtype == 2) {
        p[A[u]] = v;
        p[A[v]] = u;
        swap(A[u], A[v]);
        for (int i = 0; i < 2; ++i) {
          for (int d = 0; d <= 1; d++) {
            int a = A[i ? v : u] + d;
            if (a > 1 && a <= n) {
              update(a - 1, (p[a - 1] > p[a]) - query(a - 1, a - 1));
            }
          }
        }
      } else {
        printf("%d\n", query(u, v - 1) + 1);
      }
    }
  }
}

```