#include <bits/stdc++.h>
#pragma GCC optimize("O2")
using namespace std;
const int N = 1e6 + 10;
const long long mod = 1e9 + 7;
const long long mod2 = 998244353;
const long long inf = 8e18;
const int LOG = 22;
long long pw(long long a, long long b, long long M) {
  return (!b ? 1
             : (b & 1 ? (a * pw(a * a % M, b / 2, M)) % M
                      : pw(a * a % M, b / 2, M)));
}
int P[N], ind[N], seg[4 * N], n;
void U(int v, int tl, int tr, int p, int x) {
  if (tl == tr) {
    seg[v] = x;
    return;
  }
  int mid = (tl + tr) >> 1;
  if (p <= mid) {
    U(v << 1, tl, mid, p, x);
  } else {
    U(v << 1 | 1, mid + 1, tr, p, x);
  }
  seg[v] = seg[v << 1] + seg[v << 1 | 1];
}
int G(int v, int tl, int tr, int l, int r) {
  if (tl == l && tr == r) {
    return seg[v];
  }
  int mid = (tl + tr) >> 1, ans = 0;
  if (l <= mid) {
    ans += G(v << 1, tl, mid, l, min(r, mid));
  }
  if (r > mid) {
    ans += G(v << 1 | 1, mid + 1, tr, max(mid + 1, l), r);
  }
  return ans;
}
inline void check(int x) {
  if (x <= 0 || x > n) return;
  if (x != n && ind[x] > ind[x + 1]) {
    U(1, 1, n, x, 1);
  } else {
    U(1, 1, n, x, 0);
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> P[i];
    ind[P[i]] = i;
  }
  for (int i = 1; i < n; i++) {
    if (ind[i] > ind[i + 1]) {
      U(1, 1, n, i, 1);
    } else {
      U(1, 1, n, i, 0);
    }
  }
  int q;
  cin >> q;
  while (q--) {
    int ver;
    cin >> ver;
    if (ver == 1) {
      int l, r;
      cin >> l >> r;
      cout << 1 + G(1, 1, n, l, r) - G(1, 1, n, r, r) << "\n";
    } else {
      int i, j;
      cin >> i >> j;
      int A = P[i], B = P[j];
      swap(P[i], P[j]);
      swap(ind[P[i]], ind[P[j]]);
      check(A);
      check(A - 1);
      check(B);
      check(B - 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>
#pragma GCC optimize("O2")
using namespace std;
const int N = 1e6 + 10;
const long long mod = 1e9 + 7;
const long long mod2 = 998244353;
const long long inf = 8e18;
const int LOG = 22;
long long pw(long long a, long long b, long long M) {
  return (!b ? 1
             : (b & 1 ? (a * pw(a * a % M, b / 2, M)) % M
                      : pw(a * a % M, b / 2, M)));
}
int P[N], ind[N], seg[4 * N], n;
void U(int v, int tl, int tr, int p, int x) {
  if (tl == tr) {
    seg[v] = x;
    return;
  }
  int mid = (tl + tr) >> 1;
  if (p <= mid) {
    U(v << 1, tl, mid, p, x);
  } else {
    U(v << 1 | 1, mid + 1, tr, p, x);
  }
  seg[v] = seg[v << 1] + seg[v << 1 | 1];
}
int G(int v, int tl, int tr, int l, int r) {
  if (tl == l && tr == r) {
    return seg[v];
  }
  int mid = (tl + tr) >> 1, ans = 0;
  if (l <= mid) {
    ans += G(v << 1, tl, mid, l, min(r, mid));
  }
  if (r > mid) {
    ans += G(v << 1 | 1, mid + 1, tr, max(mid + 1, l), r);
  }
  return ans;
}
inline void check(int x) {
  if (x <= 0 || x > n) return;
  if (x != n && ind[x] > ind[x + 1]) {
    U(1, 1, n, x, 1);
  } else {
    U(1, 1, n, x, 0);
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> P[i];
    ind[P[i]] = i;
  }
  for (int i = 1; i < n; i++) {
    if (ind[i] > ind[i + 1]) {
      U(1, 1, n, i, 1);
    } else {
      U(1, 1, n, i, 0);
    }
  }
  int q;
  cin >> q;
  while (q--) {
    int ver;
    cin >> ver;
    if (ver == 1) {
      int l, r;
      cin >> l >> r;
      cout << 1 + G(1, 1, n, l, r) - G(1, 1, n, r, r) << "\n";
    } else {
      int i, j;
      cin >> i >> j;
      int A = P[i], B = P[j];
      swap(P[i], P[j]);
      swap(ind[P[i]], ind[P[j]]);
      check(A);
      check(A - 1);
      check(B);
      check(B - 1);
    }
  }
  return 0;
}

```