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>
void none(...) {}
using namespace std;
const int maxn = (int)3e5 + 5, inf = (int)1e9;
struct rmq {
  int sum;
  int lm, rm;
};
rmq t[4 * maxn];
int Next[maxn], a[maxn];
void init(int v, int l, int r) {
  t[v].lm = l, t[v].rm = r;
  if (t[v].rm - t[v].lm > 1) {
    init(v << 1, l, (r + l) >> 1);
    init((v << 1) + 1, (r + l) >> 1, r);
    t[v].sum = t[v << 1].sum + t[(v << 1) + 1].sum;
  } else {
    if (Next[l] < Next[l - 1]) {
      t[v].sum = 1;
    }
  }
}
int sum(int v, int l, int r) {
  if (l >= t[v].rm || r <= t[v].lm) return 0;
  if (l <= t[v].lm && t[v].rm <= r) {
    return t[v].sum;
  } else {
    int a = sum(v << 1, l, r);
    int b = sum((v << 1) + 1, l, r);
    return a + b;
  }
}
void change(int v, int k) {
  if (k >= t[v].rm || k < t[v].lm) return;
  if (t[v].rm - t[v].lm == 1) {
    if (Next[k] < Next[k - 1]) {
      t[v].sum = 1;
    } else
      t[v].sum = 0;
  } else {
    change(v << 1, k);
    change((v << 1) + 1, k);
    t[v].sum = t[v << 1].sum + t[(v << 1) + 1].sum;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    Next[a[i] - 1] = i;
  }
  init(1, 1, n + 1);
  int q;
  cin >> q;
  for (int it = 0; it < q; it++) {
    int op, x, y;
    cin >> op >> x >> y;
    if (op == 1) {
      int ans = sum(1, x, y);
      cout << ans + 1 << "\n";
    } else {
      swap(a[x], a[y]);
      Next[a[x] - 1] = x;
      Next[a[y] - 1] = y;
      change(1, a[y]);
      change(1, a[y] - 1);
      change(1, a[x]);
      change(1, a[x] - 1);
    }
  }
  return 0;
}
