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;
const int N = 312345;
int a[N];
int read() {
  int x;
  scanf("%d", &x);
  return x;
}
int n, d[N], ind[N];
void update(int x, int y) {
  while (x <= n) d[x] += y, x = (x | (x - 1)) + 1;
}
int sum(int x) {
  int s = 0;
  while (x > 0) s += d[x], x = (x & (x - 1));
  return s;
}
int main() {
  int m, i, j;
  n = read();
  for (i = 1; i <= n; i++) {
    a[i] = read();
    if (!ind[a[i] - 1]) update(a[i], 1);
    ind[a[i]] = i;
  }
  ind[0] = n + 1;
  m = read();
  int t, l, r, x, y;
  for (i = 1; i <= m; i++) {
    t = read();
    l = read();
    r = read();
    if (t == 1) {
      printf("%d\n", sum(r) - sum(l) + 1);
    } else {
      x = a[l];
      y = a[r];
      if (x > y) swap(x, y);
      if (ind[x] < ind[x - 1]) update(x, -1);
      if (ind[y] < ind[y - 1]) update(y, -1);
      if (y != x + 1 && ind[x + 1] < ind[x]) update(x + 1, -1);
      if (ind[y + 1] < ind[y]) update(y + 1, -1);
      ind[a[l]] = r;
      ind[a[r]] = l;
      swap(a[l], a[r]);
      if (ind[x] < ind[x - 1]) update(x, 1);
      if (ind[y] < ind[y - 1]) update(y, 1);
      if (y != x + 1 && ind[x + 1] < ind[x]) update(x + 1, 1);
      if (ind[y + 1] < ind[y]) update(y + 1, 1);
    }
  }
}
