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 mn = 300 * 1000 + 5;
int a[mn];
int b[mn];
int bad[mn];
int t[4 * mn];
void construct(int v, int l, int r) {
  if (l == r)
    t[v] = bad[r];
  else {
    int mid = (l + r) / 2;
    construct(2 * v, l, mid);
    construct(2 * v + 1, mid + 1, r);
    t[v] = t[v * 2] + t[v * 2 + 1];
  }
}
int ans(int v, int l, int r, int x, int y) {
  if (x > y) return 0;
  if (x == l && y == r) return t[v];
  int mid = (l + r) / 2;
  return ans(v * 2, l, mid, x, min(y, mid)) +
         ans(v * 2 + 1, mid + 1, r, max(x, mid + 1), y);
}
void update(int v, int l, int r, int pos, int val) {
  if (l == r)
    t[v] = val;
  else {
    int mid = (l + r) / 2;
    if (pos <= mid)
      update(v * 2, l, mid, pos, val);
    else
      update(v * 2 + 1, mid + 1, r, pos, val);
    t[v] = t[v * 2] + t[v * 2 + 1];
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    b[a[i]] = i;
  }
  for (int i = 2; i <= n; i++) {
    if (b[i] < b[i - 1]) bad[i] = 1;
  }
  construct(1, 1, n);
  int q;
  cin >> q;
  for (int k = 0; k < q; k++) {
    int t, x, y;
    cin >> t >> x >> y;
    if (t == 1) {
      cout << ans(1, 1, n, x + 1, y) + 1 << "\n";
    } else {
      int x1 = a[x], y1 = a[y];
      swap(a[x], a[y]);
      swap(b[x1], b[y1]);
      if (b[x1] < b[x1 + 1])
        update(1, 1, n, x1 + 1, 0);
      else
        update(1, 1, n, x1 + 1, 1);
      if (b[x1 - 1] < b[x1])
        update(1, 1, n, x1, 0);
      else
        update(1, 1, n, x1, 1);
      if (b[y1] < b[y1 + 1])
        update(1, 1, n, y1 + 1, 0);
      else
        update(1, 1, n, y1 + 1, 1);
      if (b[y1 - 1] < b[y1])
        update(1, 1, n, y1, 0);
      else
        update(1, 1, n, y1, 1);
    }
  }
  return 0;
}
