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 seed = 131;
const int maxn = 3e5 + 5;
const int mod = 1e9 + 7;
int n, q;
int pos[maxn];
int a[maxn];
int sum[maxn << 2];
int lazy[maxn << 2];
void push_up(int rt) { sum[rt] = sum[rt << 1] + sum[rt << 1 | 1]; }
void push_down(int rt, int len) {
  if (lazy[rt]) {
    sum[rt << 1] += lazy[rt] * (len - len / 2);
    sum[rt << 1 | 1] += lazy[rt] * (len / 2);
    lazy[rt << 1] += lazy[rt];
    lazy[rt << 1 | 1] += lazy[rt];
    lazy[rt] = 0;
  }
}
void update(int l, int r, int rt, int L, int R, int add) {
  if (L > R) return;
  if (L <= l && R >= r) {
    sum[rt] += add * (r - l + 1);
    lazy[rt] += add;
    return;
  }
  push_down(rt, r - l + 1);
  int mid = (l + r) / 2;
  if (L <= mid) update(l, mid, rt << 1, L, R, add);
  if (R > mid) update(mid + 1, r, rt << 1 | 1, L, R, add);
  push_up(rt);
}
int query(int l, int r, int rt, int p) {
  if (l == r) {
    return sum[rt];
  }
  push_down(rt, r - l + 1);
  int mid = (l + r) / 2;
  if (p <= mid)
    return query(l, mid, rt << 1, p);
  else
    return query(mid + 1, r, rt << 1 | 1, p);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    pos[a[i]] = i;
  }
  pos[n + 1] = n + 1;
  int cnt = 1;
  for (int i = 1; i <= n; i++) {
    if (pos[i] < pos[i - 1]) cnt++;
    update(1, n, 1, i, i, cnt);
  }
  scanf("%d", &q);
  while (q--) {
    int op, x, y;
    scanf("%d%d%d", &op, &x, &y);
    if (op == 1) {
      int l = query(1, n, 1, x);
      int r = query(1, n, 1, y);
      printf("%d\n", r - l + 1);
    } else {
      if (x > y) swap(x, y);
      int numx = a[x];
      int numy = a[y];
      swap(a[x], a[y]);
      swap(pos[numx], pos[numy]);
      if (numx - 1 != numy && pos[numx - 1] >= x && pos[numx - 1] <= y) {
        update(1, n, 1, numx, n, -1);
      }
      if (pos[numx + 1] >= x && pos[numx + 1] <= y) {
        update(1, n, 1, numx + 1, n, 1);
      }
      if (numy - 1 != numx && pos[numy - 1] >= x && pos[numy - 1] <= y) {
        update(1, n, 1, numy, n, 1);
      }
      if (pos[numy + 1] >= x && pos[numy + 1] <= y) {
        update(1, n, 1, numy + 1, n, -1);
      }
    }
  }
  return 0;
}
