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>
using namespace std;
int N, M, Num[333333], P, X, Y;
int D[333333], Sum[333333];
void updata(int x, int add) {
  while (x <= N) {
    Sum[x] += add;
    x += x & (-x);
  }
}
int sum(int x) {
  int ret = 0;
  while (x >= 1) {
    ret += Sum[x];
    x -= x & (-x);
  }
  return ret;
}
int main() {
  scanf("%d", &N);
  for (int n = 1; n <= N; ++n) {
    scanf("%d", Num + n);
    D[Num[n]] = n;
  }
  D[N + 1] = N + 1;
  for (int n = 1; n <= N + 1; ++n)
    if (D[n - 1] > D[n]) updata(n, 1);
  scanf("%d", &M);
  while (M--) {
    scanf("%d%d%d", &P, &X, &Y);
    if (P == 1)
      printf("%d\n", sum(Y) - sum(X) + 1);
    else {
      int n1 = Num[X], n2 = Num[Y];
      int nowx1 = D[n1]<D[n1 - 1], nowx2 = D[n1]> D[n1 + 1];
      int nowy1 = D[n2]<D[n2 - 1], nowy2 = D[n2]> D[n2 + 1];
      swap(Num[X], Num[Y]);
      swap(D[n1], D[n2]);
      int newx1 = D[n1]<D[n1 - 1], newx2 = D[n1]> D[n1 + 1];
      int newy1 = D[n2]<D[n2 - 1], newy2 = D[n2]> D[n2 + 1];
      updata(n1, newx1 - nowx1);
      if (n1 + 1 != n2) updata(n1 + 1, newx2 - nowx2);
      updata(n2, newy1 - nowy1);
      if (n2 + 1 != n1) updata(n2 + 1, newy2 - nowy2);
    }
  }
  return 0;
}
