#include <bits/stdc++.h>
using namespace std;
struct info {
  long long int x, lazy;
  info() {
    x = 0;
    lazy = 0;
  }
};
struct SegTree {
  long long int N;
  vector<info> st;
  void init(long long int n) {
    N = n;
    st.resize(4 * N + 5);
  }
  void merge(info &cur, info &l, info &r) { cur.x = l.x + r.x; }
  void Build(long long int node, long long int L, long long int R) {
    if (L == R) {
      st[node].x = 0;
      st[node].lazy = 0;
      return;
    }
    long long int MM = (L + R) / 2;
    Build(node * 2, L, MM);
    Build(node * 2 + 1, MM + 1, R);
    merge(st[node], st[node * 2], st[node * 2 + 1]);
  }
  info Query(long long int node, long long int L, long long int R,
             long long int i, long long int j) {
    if (j < L || i > R) return info();
    if (i <= L && R <= j) return st[node];
    long long int MM = (L + R) / 2;
    info left = Query(node * 2, L, MM, i, j);
    info right = Query(node * 2 + 1, MM + 1, R, i, j);
    info cur;
    merge(cur, left, right);
    return cur;
  }
  void Update(long long int node, long long int L, long long int R,
              long long int lpos, long long int rpos, long long int val) {
    if (L > rpos || R < lpos) return;
    if (L == lpos && rpos == R) {
      st[node].x = val;
      return;
    }
    long long int MM = (L + R) / 2;
    if (MM >= lpos && MM <= rpos) {
      Update(node * 2, L, MM, lpos, MM, val);
      Update(node * 2 + 1, MM + 1, R, MM + 1, rpos, val);
    } else {
      Update(node * 2, L, MM, lpos, rpos, val);
      Update(node * 2 + 1, MM + 1, R, lpos, rpos, val);
    }
    merge(st[node], st[node * 2], st[node * 2 + 1]);
  }
  info query(long long int l, long long int r) { return Query(1, 1, N, l, r); }
  void update(long long int lpos, long long int rpos, long long int val) {
    Update(1, 1, N, lpos, rpos, val);
  }
  void build() { Build(1, 1, N); }
};
void Weapons19() {
  long long int n;
  cin >> n;
  vector<long long int> v(n);
  vector<long long int> z(n + 1);
  for (long long int i = 0; i < n; i++) {
    cin >> v[i];
    z[v[i]] = i + 1;
  }
  SegTree ww;
  ww.init(n + 1);
  ww.build();
  vector<long long int> p(n + 1);
  for (long long int i = 1; i <= n; i++) {
    if (z[i] < z[i - 1]) {
      p[i] = 1;
      ww.update(i, i, 1);
    }
  }
  long long int tc;
  cin >> tc;
  while (tc--) {
    long long int a, b, c;
    cin >> a >> b >> c;
    if (a == 1) {
      if (p[b])
        cout << ww.query(b, c).x << "\n";
      else
        cout << ww.query(b, c).x + 1 << "\n";
    } else {
      long long int x = b, y = c;
      c = v[c - 1];
      b = v[b - 1];
      swap(z[c], z[b]);
      swap(v[x - 1], v[y - 1]);
      if (z[c] < z[c - 1]) {
        ww.update(c, c, 1);
        p[c] = 1;
      } else {
        ww.update(c, c, 0);
        p[c] = 0;
      }
      if (z[b] < z[b - 1]) {
        ww.update(b, b, 1);
        p[b] = 1;
      } else {
        ww.update(b, b, 0);
        p[b] = 0;
      }
      if (c != n) {
        if (z[c + 1] < z[c]) {
          ww.update(c + 1, c + 1, 1);
          p[c + 1] = 1;
        } else {
          ww.update(c + 1, c + 1, 0);
          p[c + 1] = 0;
        }
      }
      if (z[b + 1] < z[b]) {
        ww.update(b + 1, b + 1, 1);
        p[b + 1] = 1;
      } else {
        ww.update(b + 1, b + 1, 0);
        p[b + 1] = 0;
      }
    }
  }
}
int32_t main() {
  cout << fixed << setprecision(16);
  cin.sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  Weapons19();
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
struct info {
  long long int x, lazy;
  info() {
    x = 0;
    lazy = 0;
  }
};
struct SegTree {
  long long int N;
  vector<info> st;
  void init(long long int n) {
    N = n;
    st.resize(4 * N + 5);
  }
  void merge(info &cur, info &l, info &r) { cur.x = l.x + r.x; }
  void Build(long long int node, long long int L, long long int R) {
    if (L == R) {
      st[node].x = 0;
      st[node].lazy = 0;
      return;
    }
    long long int MM = (L + R) / 2;
    Build(node * 2, L, MM);
    Build(node * 2 + 1, MM + 1, R);
    merge(st[node], st[node * 2], st[node * 2 + 1]);
  }
  info Query(long long int node, long long int L, long long int R,
             long long int i, long long int j) {
    if (j < L || i > R) return info();
    if (i <= L && R <= j) return st[node];
    long long int MM = (L + R) / 2;
    info left = Query(node * 2, L, MM, i, j);
    info right = Query(node * 2 + 1, MM + 1, R, i, j);
    info cur;
    merge(cur, left, right);
    return cur;
  }
  void Update(long long int node, long long int L, long long int R,
              long long int lpos, long long int rpos, long long int val) {
    if (L > rpos || R < lpos) return;
    if (L == lpos && rpos == R) {
      st[node].x = val;
      return;
    }
    long long int MM = (L + R) / 2;
    if (MM >= lpos && MM <= rpos) {
      Update(node * 2, L, MM, lpos, MM, val);
      Update(node * 2 + 1, MM + 1, R, MM + 1, rpos, val);
    } else {
      Update(node * 2, L, MM, lpos, rpos, val);
      Update(node * 2 + 1, MM + 1, R, lpos, rpos, val);
    }
    merge(st[node], st[node * 2], st[node * 2 + 1]);
  }
  info query(long long int l, long long int r) { return Query(1, 1, N, l, r); }
  void update(long long int lpos, long long int rpos, long long int val) {
    Update(1, 1, N, lpos, rpos, val);
  }
  void build() { Build(1, 1, N); }
};
void Weapons19() {
  long long int n;
  cin >> n;
  vector<long long int> v(n);
  vector<long long int> z(n + 1);
  for (long long int i = 0; i < n; i++) {
    cin >> v[i];
    z[v[i]] = i + 1;
  }
  SegTree ww;
  ww.init(n + 1);
  ww.build();
  vector<long long int> p(n + 1);
  for (long long int i = 1; i <= n; i++) {
    if (z[i] < z[i - 1]) {
      p[i] = 1;
      ww.update(i, i, 1);
    }
  }
  long long int tc;
  cin >> tc;
  while (tc--) {
    long long int a, b, c;
    cin >> a >> b >> c;
    if (a == 1) {
      if (p[b])
        cout << ww.query(b, c).x << "\n";
      else
        cout << ww.query(b, c).x + 1 << "\n";
    } else {
      long long int x = b, y = c;
      c = v[c - 1];
      b = v[b - 1];
      swap(z[c], z[b]);
      swap(v[x - 1], v[y - 1]);
      if (z[c] < z[c - 1]) {
        ww.update(c, c, 1);
        p[c] = 1;
      } else {
        ww.update(c, c, 0);
        p[c] = 0;
      }
      if (z[b] < z[b - 1]) {
        ww.update(b, b, 1);
        p[b] = 1;
      } else {
        ww.update(b, b, 0);
        p[b] = 0;
      }
      if (c != n) {
        if (z[c + 1] < z[c]) {
          ww.update(c + 1, c + 1, 1);
          p[c + 1] = 1;
        } else {
          ww.update(c + 1, c + 1, 0);
          p[c + 1] = 0;
        }
      }
      if (z[b + 1] < z[b]) {
        ww.update(b + 1, b + 1, 1);
        p[b + 1] = 1;
      } else {
        ww.update(b + 1, b + 1, 0);
        p[b + 1] = 0;
      }
    }
  }
}
int32_t main() {
  cout << fixed << setprecision(16);
  cin.sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  Weapons19();
}

```