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, Q, S[300005], T[300005], P[300005];
bool secure[300005];
void update(int x, int v) {
  if (x == 0) return;
  for (; x <= N; x += (x & -x)) T[x] += v;
}
int query(int x) {
  if (x == 0) return 0;
  int sum = 0;
  for (; x > 0; x -= (x & -x)) sum += T[x];
  return sum;
}
void clean(int x) { update(x, -(query(x) - query(x - 1))); }
void addback(int x) {
  if (secure[x]) return;
  if (P[x] > P[x + 1]) update(x, 1);
  secure[x] = 1;
}
int main() {
  scanf("%d", &N);
  for (int i = 1; i <= N; ++i) {
    scanf("%d", &S[i]);
    P[S[i]] = i;
  }
  for (int i = 1; i <= N; ++i) addback(i);
  scanf("%d", &Q);
  int d, a, b;
  for (int z = 0; z < Q; ++z) {
    scanf("%d%d%d", &d, &a, &b);
    if (d == 1)
      printf("%d\n", query(b - 1) - query(a - 1) + 1);
    else {
      int t1 = S[a], t2 = S[b];
      clean(t1 - 1);
      clean(t1);
      clean(t2 - 1);
      clean(t2);
      secure[t1 - 1] = secure[t1] = secure[t2 - 1] = secure[t2] = 0;
      swap(S[a], S[b]);
      P[t1] = b;
      P[t2] = a;
      addback(t1 - 1);
      addback(t1);
      addback(t2 - 1);
      addback(t2);
    }
  }
}
