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;
int a[300005];
int rev[300005];
int der[1200005];
void makeT(int idx, int l, int r) {
  if (l == r) {
    der[idx] = (l > 0 ? (rev[l] < rev[l - 1]) : 0);
    return;
  }
  int m = (l + r) / 2;
  makeT(idx * 2, l, m);
  makeT(idx * 2 + 1, m + 1, r);
  der[idx] = der[idx * 2] + der[idx * 2 + 1];
}
int getSum(int idx, int l, int r, int a, int b) {
  if (l == a && r == b) return der[idx];
  int m = (a + b) / 2;
  int ans = 0;
  if (l <= m) ans += getSum(idx * 2, l, min(r, m), a, m);
  if (r >= m + 1) ans += getSum(idx * 2 + 1, max(l, m + 1), r, m + 1, b);
  return ans;
}
void setT(int idx, int pos, int val, int a, int b) {
  if (a == b) {
    der[idx] = val;
    return;
  }
  int m = (a + b) / 2;
  if (pos <= m)
    setT(idx * 2, pos, val, a, m);
  else
    setT(idx * 2 + 1, pos, val, m + 1, b);
  der[idx] = der[idx * 2] + der[idx * 2 + 1];
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    rev[a[i]] = i;
  }
  int q;
  scanf("%d", &q);
  makeT(1, 1, n);
  for (int t = 0; t < q; t++) {
    int tp, x, y;
    scanf("%d%d%d", &tp, &x, &y);
    if (tp == 1) {
      int ans = 1 + (x + 1 <= y ? getSum(1, x + 1, y, 1, n) : 0);
      printf("%d\n", ans);
    } else {
      swap(a[x], a[y]);
      rev[a[x]] = x;
      rev[a[y]] = y;
      if (a[x] > 1) setT(1, a[x], rev[a[x]] < rev[a[x] - 1], 1, n);
      if (a[x] + 1 <= n) setT(1, a[x] + 1, rev[a[x] + 1] < rev[a[x]], 1, n);
      if (a[y] > 1) setT(1, a[y], rev[a[y]] < rev[a[y] - 1], 1, n);
      if (a[y] + 1 <= n) setT(1, a[y] + 1, rev[a[y] + 1] < rev[a[y]], 1, n);
    }
  }
  return 0;
}
