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
#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:16777216")
using namespace std;
long N, Q;
long A[1000007 + 7];
long BIT[1000007 + 7];
void Update(long i, long v) {
  while (i <= 1000007) {
    BIT[i] += v;
    i += i & -i;
  }
}
long Find(long i) {
  long s = 0;
  while (i) {
    s += BIT[i];
    i -= i & -i;
  }
  return s;
}
int main(void) {
  long i, j, t, Icase, k = 0;
  cin >> N;
  A[0] = 0;
  A[N + 1] = 7777777;
  map<long, long> Pos;
  Pos[0] = 0;
  Pos[N + 1] = N + 1;
  for (i = 1; i <= N; i++) {
    cin >> A[i];
    Pos[A[i]] = i;
  }
  for (i = 1; i <= N; i++) {
    if (Pos[i - 1] > Pos[i]) Update(i, 1);
  }
  cin >> Q;
  while (Q--) {
    scanf("%ld%ld%ld", &t, &i, &j);
    if (t == 1) {
      long Ans = Find(j) - Find(i);
      printf("%ld\n", Ans + 1);
    } else if (t == 2) {
      long v1 = A[i];
      long v2 = A[j];
      if (v1 > v2) swap(v1, v2);
      if (Pos[v1 - 1] > Pos[v1]) Update(v1, -1);
      if (Pos[v1] > Pos[v1 + 1]) Update(v1 + 1, -1);
      if (v1 + 1 != v2 && Pos[v2 - 1] > Pos[v2]) Update(v2, -1);
      if (Pos[v2] > Pos[v2 + 1]) Update(v2 + 1, -1);
      swap(A[i], A[j]);
      Pos[A[i]] = i;
      Pos[A[j]] = j;
      if (Pos[v1 - 1] > Pos[v1]) Update(v1, 1);
      if (Pos[v1] > Pos[v1 + 1]) Update(v1 + 1, 1);
      if (v1 + 1 != v2 && Pos[v2 - 1] > Pos[v2]) Update(v2, 1);
      if (Pos[v2] > Pos[v2 + 1]) Update(v2 + 1, 1);
    }
  }
  return 0;
}
