#include <bits/stdc++.h>
using namespace std;
template <typename T>
T nextInt() {
  T x = 0, p = 1;
  char ch;
  do {
    ch = getchar();
  } while (ch <= ' ');
  if (ch == '-') {
    p = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + (ch - '0');
    ch = getchar();
  }
  return x * p;
}
const int maxN = (int)3e5 + 10;
const int maxK = 17;
const int INF = (int)1e9;
const int mod = (int)1e9 + 7;
const long long LLINF = (long long)1e18;
int n, q;
int a[maxN];
int p[maxN];
template <typename T>
T myFunction(T a, T b) {
  return a + b;
}
template <class T>
struct segmentTree {
  int n;
  T *t;
  segmentTree(int n) : n(n) {
    t = new T[4 * n];
    fill(t, t + 4 * n, 0);
    build(1, 0, n - 1);
  }
  void build(int v, int l, int r) {
    if (l == r) {
      t[v] = 0;
    } else {
      int m = (l + r) / 2;
      build(v + v, l, m);
      build(v + v + 1, m + 1, r);
      t[v] = t[v + v] + t[v + 1 + v] + (p[m] > p[m + 1]);
    }
  }
  T get(int l, int r) { return get(1, l, r, 0, n - 1); }
  T get(int v, int l, int r, int tl, int tr) {
    if (tl >= l && tr <= r) {
      return t[v];
    }
    int tm = (tl + tr) / 2;
    if (l > tm) {
      return get(v + v + 1, l, r, tm + 1, tr);
    }
    if (r <= tm) {
      return get(v + v, l, r, tl, tm);
    }
    return myFunction(get(v + v, l, r, tl, tm),
                      get(v + v + 1, l, r, tm + 1, tr)) +
           (p[tm] > p[tm + 1]);
  }
  void modifyElement(int pos) { modifyElement(1, pos, 0, n - 1); }
  void modifyElement(int v, int pos, int tl, int tr) {
    if (tl == tr) {
      return;
    }
    int tm = (tl + tr) / 2;
    if (pos <= tm) {
      modifyElement(v + v, pos, tl, tm);
    } else {
      modifyElement(v + v + 1, pos, tm + 1, tr);
    }
    t[v] = myFunction(t[v + v], t[v + v + 1]) + (p[tm] > p[tm + 1]);
  }
};
int main() {
  n = nextInt<int>();
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    a[i]--;
    p[a[i]] = i;
  }
  segmentTree<int> t(n);
  q = nextInt<int>();
  while (q--) {
    int type, l, r;
    type = nextInt<int>();
    l = nextInt<int>() - 1;
    r = nextInt<int>() - 1;
    if (type == 1) {
      printf("%d\n", t.get(l, r) + 1);
    } else {
      swap(a[l], a[r]);
      swap(p[a[l]], p[a[r]]);
      t.modifyElement(a[l]);
      t.modifyElement(a[r]);
    }
  }
  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;
template <typename T>
T nextInt() {
  T x = 0, p = 1;
  char ch;
  do {
    ch = getchar();
  } while (ch <= ' ');
  if (ch == '-') {
    p = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + (ch - '0');
    ch = getchar();
  }
  return x * p;
}
const int maxN = (int)3e5 + 10;
const int maxK = 17;
const int INF = (int)1e9;
const int mod = (int)1e9 + 7;
const long long LLINF = (long long)1e18;
int n, q;
int a[maxN];
int p[maxN];
template <typename T>
T myFunction(T a, T b) {
  return a + b;
}
template <class T>
struct segmentTree {
  int n;
  T *t;
  segmentTree(int n) : n(n) {
    t = new T[4 * n];
    fill(t, t + 4 * n, 0);
    build(1, 0, n - 1);
  }
  void build(int v, int l, int r) {
    if (l == r) {
      t[v] = 0;
    } else {
      int m = (l + r) / 2;
      build(v + v, l, m);
      build(v + v + 1, m + 1, r);
      t[v] = t[v + v] + t[v + 1 + v] + (p[m] > p[m + 1]);
    }
  }
  T get(int l, int r) { return get(1, l, r, 0, n - 1); }
  T get(int v, int l, int r, int tl, int tr) {
    if (tl >= l && tr <= r) {
      return t[v];
    }
    int tm = (tl + tr) / 2;
    if (l > tm) {
      return get(v + v + 1, l, r, tm + 1, tr);
    }
    if (r <= tm) {
      return get(v + v, l, r, tl, tm);
    }
    return myFunction(get(v + v, l, r, tl, tm),
                      get(v + v + 1, l, r, tm + 1, tr)) +
           (p[tm] > p[tm + 1]);
  }
  void modifyElement(int pos) { modifyElement(1, pos, 0, n - 1); }
  void modifyElement(int v, int pos, int tl, int tr) {
    if (tl == tr) {
      return;
    }
    int tm = (tl + tr) / 2;
    if (pos <= tm) {
      modifyElement(v + v, pos, tl, tm);
    } else {
      modifyElement(v + v + 1, pos, tm + 1, tr);
    }
    t[v] = myFunction(t[v + v], t[v + v + 1]) + (p[tm] > p[tm + 1]);
  }
};
int main() {
  n = nextInt<int>();
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    a[i]--;
    p[a[i]] = i;
  }
  segmentTree<int> t(n);
  q = nextInt<int>();
  while (q--) {
    int type, l, r;
    type = nextInt<int>();
    l = nextInt<int>() - 1;
    r = nextInt<int>() - 1;
    if (type == 1) {
      printf("%d\n", t.get(l, r) + 1);
    } else {
      swap(a[l], a[r]);
      swap(p[a[l]], p[a[r]]);
      t.modifyElement(a[l]);
      t.modifyElement(a[r]);
    }
  }
  return 0;
}

```