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;
const double pi = 2.0 * acos(-1.0);
const int inf = (int)1e9;
const int S = 600;
int mas[300000], p[300000];
int cl[300000], add[600];
int n;
void addto(int l, int r, int num) {
  int cnt = l;
  for (int i = l; i <= r; i++, cnt++) {
    if (cnt % S == 0) break;
    cl[cnt] += num;
  }
  if (cnt > r) return;
  while (cnt + S <= r) {
    add[cnt / S] += num;
    cnt += S;
  }
  while (cnt <= r) {
    cl[cnt] += num;
    cnt++;
  }
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> mas[i];
    mas[i]--;
    p[mas[i]] = i;
  }
  cl[0] = 1;
  for (int i = 1; i < n; i++) {
    if (p[i] < p[i - 1])
      cl[i] = cl[i - 1] + 1;
    else
      cl[i] = cl[i - 1];
  }
  int q;
  cin >> q;
  for (int i = 0; i < q; i++) {
    int a, b, c;
    scanf("%d%d%d", &a, &b, &c);
    b--, c--;
    if (a == 2) {
      int f = mas[b], s = mas[c];
      swap(mas[b], mas[c]);
      p[f] = c, p[s] = b;
      if (f > 0 && p[f - 1] > p[f] &&
          cl[f] + add[f / S] == cl[f - 1] + add[(f - 1) / S])
        addto(f, n - 1, 1);
      if (f < n - 1 && p[f + 1] < p[f] &&
          cl[f] + add[f / S] == cl[f + 1] + add[(f + 1) / S])
        addto(f + 1, n - 1, 1);
      if (f > 0 && p[f - 1] < p[f] &&
          cl[f] + add[f / S] != cl[f - 1] + add[(f - 1) / S])
        addto(f, n - 1, -1);
      if (f < n - 1 && p[f + 1] > p[f] &&
          cl[f] + add[f / S] != cl[f + 1] + add[(f + 1) / S])
        addto(f + 1, n - 1, -1);
      if (s > 0 && p[s - 1] > p[s] &&
          cl[s] + add[s / S] == cl[s - 1] + add[(s - 1) / S])
        addto(s, n - 1, 1);
      if (s < n - 1 && p[s + 1] < p[s] &&
          cl[s] + add[s / S] == cl[s + 1] + add[(s + 1) / S])
        addto(s + 1, n - 1, 1);
      if (s > 0 && p[s - 1] < p[s] &&
          cl[s] + add[s / S] != cl[s - 1] + add[(s - 1) / S])
        addto(s, n - 1, -1);
      if (s < n - 1 && p[s + 1] > p[s] &&
          cl[s] + add[s / S] != cl[s + 1] + add[(s + 1) / S])
        addto(s + 1, n - 1, -1);
    } else {
      int clr = cl[c] + add[c / S];
      int cll = cl[b] + add[b / S];
      printf("%d\n", clr - cll + 1);
    }
  }
  return 0;
}
