#include <bits/stdc++.h>
using namespace std;
const int INF = 1001001001;
const long long INFLL = 1001001001001001001LL;
template <typename T>
void pv(T a, T b) {
  for (T i = a; i != b; ++i) cout << *i << " ";
  cout << endl;
}
template <typename T>
void chmin(T& a, T b) {
  if (a > b) a = b;
}
template <typename T>
void chmax(T& a, T b) {
  if (a < b) a = b;
}
int n;
int A[300050], P[300050];
int B[300050], J[300050];
void add(int x, int v) {
  while (x <= 300000) {
    B[x] += v;
    x += x & -x;
  }
}
int sum(int x) {
  int s = 0;
  while (x > 0) {
    s += B[x];
    x -= x & -x;
  }
  return s;
}
void check(int u) {
  if (u == 1 || u > n) return;
  if (P[u] < P[u - 1] && J[u] == 0) {
    add(u, 1);
    J[u] = 1;
  } else if (P[u] > P[u - 1] && J[u] == 1) {
    add(u, -1);
    J[u] = 0;
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> A[i];
    P[A[i]] = i;
  }
  for (int i = 2; i <= n; ++i) {
    if (P[i] < P[i - 1]) {
      add(i, 1);
      J[i] = 1;
    }
  }
  int q;
  cin >> q;
  for (int i = 0; i < q; ++i) {
    int p, x, y;
    cin >> p >> x >> y;
    if (p == 1) {
      cout << sum(y) - sum(x) + 1 << endl;
    } else {
      --x;
      --y;
      swap(A[x], A[y]);
      swap(P[A[x]], P[A[y]]);
      check(A[x]);
      check(A[x] + 1);
      check(A[y]);
      check(A[y] + 1);
    }
  }
  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>
using namespace std;
const int INF = 1001001001;
const long long INFLL = 1001001001001001001LL;
template <typename T>
void pv(T a, T b) {
  for (T i = a; i != b; ++i) cout << *i << " ";
  cout << endl;
}
template <typename T>
void chmin(T& a, T b) {
  if (a > b) a = b;
}
template <typename T>
void chmax(T& a, T b) {
  if (a < b) a = b;
}
int n;
int A[300050], P[300050];
int B[300050], J[300050];
void add(int x, int v) {
  while (x <= 300000) {
    B[x] += v;
    x += x & -x;
  }
}
int sum(int x) {
  int s = 0;
  while (x > 0) {
    s += B[x];
    x -= x & -x;
  }
  return s;
}
void check(int u) {
  if (u == 1 || u > n) return;
  if (P[u] < P[u - 1] && J[u] == 0) {
    add(u, 1);
    J[u] = 1;
  } else if (P[u] > P[u - 1] && J[u] == 1) {
    add(u, -1);
    J[u] = 0;
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> A[i];
    P[A[i]] = i;
  }
  for (int i = 2; i <= n; ++i) {
    if (P[i] < P[i - 1]) {
      add(i, 1);
      J[i] = 1;
    }
  }
  int q;
  cin >> q;
  for (int i = 0; i < q; ++i) {
    int p, x, y;
    cin >> p >> x >> y;
    if (p == 1) {
      cout << sum(y) - sum(x) + 1 << endl;
    } else {
      --x;
      --y;
      swap(A[x], A[y]);
      swap(P[A[x]], P[A[y]]);
      check(A[x]);
      check(A[x] + 1);
      check(A[y]);
      check(A[y] + 1);
    }
  }
  return 0;
}

```