#include <bits/stdc++.h>
using namespace std;
const long long int MOD = 1000000007;
const long long BIG = 1446803456761533460LL;
const int Big = 336860180;
const long long int INF = LONG_LONG_MAX;
const long long int adj4[4][2] = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
const long long int adj8[8][2] = {{0, 1},   {0, -1}, {1, 0},  {-1, 0},
                                  {-1, -1}, {1, -1}, {-1, 1}, {1, 1}};
long long int clk_ar[10];
long long int gcd(long long int a, long long int b) {
  return b == 0 ? a : gcd(b, a % b);
}
long long int powMod(long long int a, long long int b) {
  long long int n = 1;
  long long int p = a;
  while (b > 0) {
    if (b % 2 == 1) {
      n *= p;
      n %= MOD;
    }
    p *= p;
    p %= MOD;
    b /= 2;
  }
  return n;
}
long long int modularInverse(long long int a) { return powMod(a, MOD - 2); }
stringstream sss;
const long long int maxn = 300010;
long long int ar[maxn];
long long int id[maxn];
long long int fen[maxn];
void update(long long int i, long long int v) {
  for (++i; i < maxn; i += i & (-i)) fen[i] += v;
}
long long int get(long long int i) {
  long long int s = 0;
  for (++i; i > 0; i -= i & (-i)) s += fen[i];
  return s;
}
void MAIN() {
  long long int n;
  cin >> n;
  for (long long int i = 0; i < (n); ++i) {
    long long int x;
    cin >> x;
    --x;
    ar[i] = x;
    id[x] = i;
  }
  for (long long int i = 0; i < (n - 1); ++i) {
    if (id[i] < id[i + 1]) {
      update(i, 1);
    }
  }
  long long int Q;
  cin >> Q;
  for (long long int q = 0; q < (Q); ++q) {
    long long int t, xi, yi;
    cin >> t >> xi >> yi;
    --xi, --yi;
    if (t == 1) {
      long long int x = xi;
      long long int y = yi;
      long long int s = get(y - 1) - get(x - 1);
      cout << (y - x + 1 - s) << '\n';
    } else if (t == 2) {
      long long int x = ar[xi];
      long long int y = ar[yi];
      if (x > y) swap(x, y);
      if (x > 0) {
        if (id[x - 1] < id[x]) {
          update(x - 1, -1);
        }
        if (id[x - 1] < id[y]) {
          update(x - 1, 1);
        }
      }
      if (y + 1 < n) {
        if (id[y] < id[y + 1]) {
          update(y, -1);
        }
        if (id[x] < id[y + 1]) {
          update(y, 1);
        }
      }
      if (x == y - 1) {
        if (id[x] < id[y]) {
          update(x, -1);
        } else {
          update(x, 1);
        }
      } else {
        if (id[x] < id[x + 1]) {
          update(x, -1);
        }
        if (id[y] < id[x + 1]) {
          update(x, 1);
        }
      }
      if (y - 1 != x) {
        if (id[y - 1] < id[y]) {
          update(y - 1, -1);
        }
        if (id[y - 1] < id[x]) {
          update(y - 1, 1);
        }
      }
      swap(ar[xi], ar[yi]);
      swap(id[x], id[y]);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cout << fixed << setprecision(10);
  sss << R"(
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
    )";
  MAIN();
  return 0;
}

### Prompt

Create a solution in Cpp for 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 long long int MOD = 1000000007;
const long long BIG = 1446803456761533460LL;
const int Big = 336860180;
const long long int INF = LONG_LONG_MAX;
const long long int adj4[4][2] = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
const long long int adj8[8][2] = {{0, 1},   {0, -1}, {1, 0},  {-1, 0},
                                  {-1, -1}, {1, -1}, {-1, 1}, {1, 1}};
long long int clk_ar[10];
long long int gcd(long long int a, long long int b) {
  return b == 0 ? a : gcd(b, a % b);
}
long long int powMod(long long int a, long long int b) {
  long long int n = 1;
  long long int p = a;
  while (b > 0) {
    if (b % 2 == 1) {
      n *= p;
      n %= MOD;
    }
    p *= p;
    p %= MOD;
    b /= 2;
  }
  return n;
}
long long int modularInverse(long long int a) { return powMod(a, MOD - 2); }
stringstream sss;
const long long int maxn = 300010;
long long int ar[maxn];
long long int id[maxn];
long long int fen[maxn];
void update(long long int i, long long int v) {
  for (++i; i < maxn; i += i & (-i)) fen[i] += v;
}
long long int get(long long int i) {
  long long int s = 0;
  for (++i; i > 0; i -= i & (-i)) s += fen[i];
  return s;
}
void MAIN() {
  long long int n;
  cin >> n;
  for (long long int i = 0; i < (n); ++i) {
    long long int x;
    cin >> x;
    --x;
    ar[i] = x;
    id[x] = i;
  }
  for (long long int i = 0; i < (n - 1); ++i) {
    if (id[i] < id[i + 1]) {
      update(i, 1);
    }
  }
  long long int Q;
  cin >> Q;
  for (long long int q = 0; q < (Q); ++q) {
    long long int t, xi, yi;
    cin >> t >> xi >> yi;
    --xi, --yi;
    if (t == 1) {
      long long int x = xi;
      long long int y = yi;
      long long int s = get(y - 1) - get(x - 1);
      cout << (y - x + 1 - s) << '\n';
    } else if (t == 2) {
      long long int x = ar[xi];
      long long int y = ar[yi];
      if (x > y) swap(x, y);
      if (x > 0) {
        if (id[x - 1] < id[x]) {
          update(x - 1, -1);
        }
        if (id[x - 1] < id[y]) {
          update(x - 1, 1);
        }
      }
      if (y + 1 < n) {
        if (id[y] < id[y + 1]) {
          update(y, -1);
        }
        if (id[x] < id[y + 1]) {
          update(y, 1);
        }
      }
      if (x == y - 1) {
        if (id[x] < id[y]) {
          update(x, -1);
        } else {
          update(x, 1);
        }
      } else {
        if (id[x] < id[x + 1]) {
          update(x, -1);
        }
        if (id[y] < id[x + 1]) {
          update(x, 1);
        }
      }
      if (y - 1 != x) {
        if (id[y - 1] < id[y]) {
          update(y - 1, -1);
        }
        if (id[y - 1] < id[x]) {
          update(y - 1, 1);
        }
      }
      swap(ar[xi], ar[yi]);
      swap(id[x], id[y]);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cout << fixed << setprecision(10);
  sss << R"(
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
    )";
  MAIN();
  return 0;
}

```