#include <bits/stdc++.h>
using namespace std;
template <class T>
inline T bigmod(T p, T e, T M) {
  if (e == 0) return 1;
  if (e % 2 == 0) {
    long long t = bigmod(p, e / 2, M);
    return (T)((t * t) % M);
  }
  return (T)((long long)bigmod(p, e - 1, M) * (long long)p) % M;
}
template <class T>
inline T gcd(T a, T b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
template <class T>
inline T modinverse(T a, T M) {
  return bigmod(a, M - 2, M);
}
int pr[400001];
int ar[400001];
int bit[400001];
int nxt[400001];
map<long long, long long> mm;
set<long long> st;
void insert(int ii, int v) {
  while (ii <= 400000) {
    bit[ii] += v;
    ii += ii & -ii;
  }
}
int query(int ii) {
  int ret = 0;
  while (ii > 0) {
    ret += bit[ii];
    ii -= ii & -ii;
  }
  return ret;
}
int main() {
  long long a, b = 0, c, d, e, f, g, h, x, y, z;
  cin >> a;
  for (int i = (1); i < (a + 1); ++i) {
    cin >> ar[i];
    mm[ar[i]] = i;
  }
  insert(1, 1);
  for (int i = (2); i < (a + 1); ++i) {
    if (mm[i] > mm[i - 1])
      pr[i] = i - 1, nxt[i - 1] = i;
    else
      insert(i, 1);
  }
  cin >> b;
  for (int i = (0); i < (b); ++i) {
    cin >> c >> d >> e;
    if (c == 1) {
      if (d > e) swap(d, e);
      cout << query(e) - query(d) + 1 << endl;
    } else {
      mm[ar[d]] = e;
      mm[ar[e]] = d;
      if (d > e) swap(d, e);
      if (nxt[ar[d]] != 0 && mm[nxt[ar[d]]] < mm[ar[d]]) {
        insert(nxt[ar[d]], 1);
        pr[nxt[ar[d]]] = 0;
        nxt[ar[d]] = 0;
      }
      if (pr[ar[d]] == 0 && ar[d] != 1 && mm[ar[d] - 1] < mm[ar[d]]) {
        insert(ar[d], -1);
        pr[ar[d]] = ar[d] - 1;
        nxt[ar[d] - 1] = ar[d];
      }
      if (pr[ar[e]] != 0 && mm[pr[ar[e]]] > mm[ar[e]]) {
        insert(ar[e], 1);
        nxt[pr[ar[e]]] = 0;
        pr[ar[e]] = 0;
      }
      if (nxt[ar[e]] == 0 && ar[e] != a && mm[ar[e] + 1] > mm[ar[e]]) {
        insert(ar[e] + 1, -1);
        nxt[ar[e]] = ar[e] + 1;
        pr[ar[e] + 1] = ar[e];
      }
      swap(ar[d], ar[e]);
    }
  }
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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 <class T>
inline T bigmod(T p, T e, T M) {
  if (e == 0) return 1;
  if (e % 2 == 0) {
    long long t = bigmod(p, e / 2, M);
    return (T)((t * t) % M);
  }
  return (T)((long long)bigmod(p, e - 1, M) * (long long)p) % M;
}
template <class T>
inline T gcd(T a, T b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
template <class T>
inline T modinverse(T a, T M) {
  return bigmod(a, M - 2, M);
}
int pr[400001];
int ar[400001];
int bit[400001];
int nxt[400001];
map<long long, long long> mm;
set<long long> st;
void insert(int ii, int v) {
  while (ii <= 400000) {
    bit[ii] += v;
    ii += ii & -ii;
  }
}
int query(int ii) {
  int ret = 0;
  while (ii > 0) {
    ret += bit[ii];
    ii -= ii & -ii;
  }
  return ret;
}
int main() {
  long long a, b = 0, c, d, e, f, g, h, x, y, z;
  cin >> a;
  for (int i = (1); i < (a + 1); ++i) {
    cin >> ar[i];
    mm[ar[i]] = i;
  }
  insert(1, 1);
  for (int i = (2); i < (a + 1); ++i) {
    if (mm[i] > mm[i - 1])
      pr[i] = i - 1, nxt[i - 1] = i;
    else
      insert(i, 1);
  }
  cin >> b;
  for (int i = (0); i < (b); ++i) {
    cin >> c >> d >> e;
    if (c == 1) {
      if (d > e) swap(d, e);
      cout << query(e) - query(d) + 1 << endl;
    } else {
      mm[ar[d]] = e;
      mm[ar[e]] = d;
      if (d > e) swap(d, e);
      if (nxt[ar[d]] != 0 && mm[nxt[ar[d]]] < mm[ar[d]]) {
        insert(nxt[ar[d]], 1);
        pr[nxt[ar[d]]] = 0;
        nxt[ar[d]] = 0;
      }
      if (pr[ar[d]] == 0 && ar[d] != 1 && mm[ar[d] - 1] < mm[ar[d]]) {
        insert(ar[d], -1);
        pr[ar[d]] = ar[d] - 1;
        nxt[ar[d] - 1] = ar[d];
      }
      if (pr[ar[e]] != 0 && mm[pr[ar[e]]] > mm[ar[e]]) {
        insert(ar[e], 1);
        nxt[pr[ar[e]]] = 0;
        pr[ar[e]] = 0;
      }
      if (nxt[ar[e]] == 0 && ar[e] != a && mm[ar[e] + 1] > mm[ar[e]]) {
        insert(ar[e] + 1, -1);
        nxt[ar[e]] = ar[e] + 1;
        pr[ar[e] + 1] = ar[e];
      }
      swap(ar[d], ar[e]);
    }
  }
}

```