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 *a = new int[1048576];
int l(int q) {
  int w = 1, r = 0, t;
  for (t = 262144; t; t >>= 1) {
    w <<= 1;
    if (q >= t) {
      q -= t;
      r += a[w];
      w++;
    }
  }
  return r;
}
void k(int w, int e) {
  for (w += 524288; w; w >>= 1) a[w] += e;
}
int main() {
  int q, w, e, r, t, c, v, s[300000], d[4], *f = new int[300000];
  for (w = 0; w < 1048576; w++) a[w] = 0;
  cin >> q;
  for (w = 0; w < q; w++) {
    scanf("%d", &f[w]);
    f[w]--;
    s[f[w]] = w;
  }
  for (w = 0; w < q - 1; w++)
    if (s[w] > s[w + 1]) k(w, 1);
  for (cin >> w; w; w--) {
    scanf("%d%d%d", &t, &e, &r);
    e--;
    r--;
    if (t == 1)
      printf("%d\n", l(r) - l(e) + 1);
    else {
      if (f[e] > f[r]) {
        t = e;
        e = r;
        r = t;
      }
      t = 0;
      if (f[e] > 0) {
        d[t] = f[e] - 1;
        t++;
      }
      d[t] = f[e];
      t++;
      if (f[e] < f[r] - 1) {
        d[t] = f[r] - 1;
        t++;
      }
      if (f[r] < q - 1) {
        d[t] = f[r];
        t++;
      }
      for (c = 0; c < t; c++)
        if (s[d[c]] > s[d[c] + 1]) k(d[c], -1);
      c = f[e];
      f[e] = f[r];
      f[r] = c;
      s[f[e]] = e;
      s[f[r]] = r;
      for (c = 0; c < t; c++)
        if (s[d[c]] > s[d[c] + 1]) k(d[c], 1);
    }
  }
  return 0;
}
