#include <bits/stdc++.h>
using namespace ::std;
const int maxn = 3e5 + 500;
const int mod = 1e9 + 7;
const int inf = 1e9 + 500;
int p[maxn];
int q[maxn];
struct no {
  int ans, sar, tah;
};
no seg[maxn * 4];
int tohi;
no toohi;
no merg(no a, no b) {
  if (a.ans == tohi) {
    return b;
  }
  if (b.ans == tohi) {
    return a;
  }
  no c;
  c.sar = a.sar;
  c.tah = b.tah;
  c.ans = a.ans + b.ans;
  if (a.tah < b.sar) {
    c.ans--;
  }
  return c;
}
void update(int L, int R, int node, int x, int v) {
  if (L + 1 == R) {
    seg[node].ans = 1;
    seg[node].sar = v;
    seg[node].tah = v;
    return;
  }
  int mid = (L + R) / 2;
  if (x < mid) {
    update(L, mid, 2 * node, x, v);
  } else {
    update(mid, R, 2 * node + 1, x, v);
  }
  seg[node] = merg(seg[2 * node], seg[2 * node + 1]);
}
no finds(int L, int R, int node, int l, int r) {
  if (l <= L && R <= r) {
    return seg[node];
  }
  if (R <= l || r <= L) {
    return toohi;
  }
  int mid = (L + R) / 2;
  return merg(finds(L, mid, 2 * node, l, r), finds(mid, R, 2 * node + 1, l, r));
}
void pri(no k) {
  cout << "ans :" << k.ans << "  sar :" << k.sar << "  tah :" << k.tah << endl;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  tohi = -1e7;
  toohi.ans = tohi;
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> p[i];
    update(1, n + 1, 1, p[i], i);
  }
  int qq;
  cin >> qq;
  for (int qw = 0; qw < qq; qw++) {
    int t, l, r;
    cin >> t >> l >> r;
    if (t == 1) {
      cout << finds(1, n + 1, 1, l, r + 1).ans << endl;
    } else {
      swap(p[l], p[r]);
      update(1, n + 1, 1, p[l], l);
      update(1, n + 1, 1, p[r], r);
    }
  }
}

### 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;
const int maxn = 3e5 + 500;
const int mod = 1e9 + 7;
const int inf = 1e9 + 500;
int p[maxn];
int q[maxn];
struct no {
  int ans, sar, tah;
};
no seg[maxn * 4];
int tohi;
no toohi;
no merg(no a, no b) {
  if (a.ans == tohi) {
    return b;
  }
  if (b.ans == tohi) {
    return a;
  }
  no c;
  c.sar = a.sar;
  c.tah = b.tah;
  c.ans = a.ans + b.ans;
  if (a.tah < b.sar) {
    c.ans--;
  }
  return c;
}
void update(int L, int R, int node, int x, int v) {
  if (L + 1 == R) {
    seg[node].ans = 1;
    seg[node].sar = v;
    seg[node].tah = v;
    return;
  }
  int mid = (L + R) / 2;
  if (x < mid) {
    update(L, mid, 2 * node, x, v);
  } else {
    update(mid, R, 2 * node + 1, x, v);
  }
  seg[node] = merg(seg[2 * node], seg[2 * node + 1]);
}
no finds(int L, int R, int node, int l, int r) {
  if (l <= L && R <= r) {
    return seg[node];
  }
  if (R <= l || r <= L) {
    return toohi;
  }
  int mid = (L + R) / 2;
  return merg(finds(L, mid, 2 * node, l, r), finds(mid, R, 2 * node + 1, l, r));
}
void pri(no k) {
  cout << "ans :" << k.ans << "  sar :" << k.sar << "  tah :" << k.tah << endl;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  tohi = -1e7;
  toohi.ans = tohi;
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> p[i];
    update(1, n + 1, 1, p[i], i);
  }
  int qq;
  cin >> qq;
  for (int qw = 0; qw < qq; qw++) {
    int t, l, r;
    cin >> t >> l >> r;
    if (t == 1) {
      cout << finds(1, n + 1, 1, l, r + 1).ans << endl;
    } else {
      swap(p[l], p[r]);
      update(1, n + 1, 1, p[l], l);
      update(1, n + 1, 1, p[r], r);
    }
  }
}

```