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>
const int MaxN = 3e5 + 7;
int n, m, a[MaxN], p[MaxN];
int leaf, tr[MaxN * 2];
void build(int n) {
  leaf = n;
  for (int i = 0; i < n; ++i) tr[i + leaf] = a[i] > a[i + 1];
  for (int i = leaf - 1; i; --i) tr[i] = tr[i << 1] + tr[i << 1 | 1];
}
void upd(int x, int val) {
  tr[x += leaf] += val;
  while (x >>= 1) tr[x] += val;
}
int find(int l, int r) {
  int ans = 0;
  for (l += leaf, r += leaf; r && l <= r; l >>= 1, r >>= 1) {
    if (l & 1) ans += tr[l++];
    if (~r & 1) ans += tr[r--];
  }
  return ans;
}
void adjust(int x) {
  int _x = x + leaf;
  if (x) {
    if (tr[_x - 1] && a[x - 1] < a[x]) upd(x - 1, -1);
    if (!tr[_x - 1] && a[x - 1] > a[x]) upd(x - 1, 1);
  }
  if (x < n - 1) {
    if (tr[_x] && a[x] < a[x + 1]) upd(x, -1);
    if (!tr[_x] && a[x] > a[x + 1]) upd(x, 1);
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%d", p + i);
    a[--p[i]] = i;
  }
  build(n);
  scanf("%d", &m);
  for (int i = 0; i < m; ++i) {
    int type;
    scanf("%d", &type);
    if (type & 1) {
      int l, r;
      scanf("%d %d", &l, &r);
      printf("%d\n", 1 + find(l - 1, r - 2));
    } else {
      int x, y;
      scanf("%d %d", &x, &y);
      --x;
      --y;
      std::swap(a[p[x]], a[p[y]]);
      std::swap(p[x], p[y]);
      adjust(p[x]);
      adjust(p[y]);
    }
  }
  return 0;
}
