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;
#pragma comment(linker, "/STACK:64000000")
const int MAXN = 300003;
int n, m, cmd, l, r;
int a[MAXN];
int pos[MAXN];
int f[MAXN];
int flags[MAXN];
int FLAG;
void add(int x, int d) {
  while (x < n) {
    f[x] += d;
    x |= (x + 1);
  }
}
int get(int x) {
  int res = 0;
  while (x >= 0) {
    res += f[x];
    x = (x & (x + 1)) - 1;
  }
  return res;
}
int get(int l, int r) { return get(r) - get(l - 1); }
inline void check(int x, int d) {
  if (x < 1 || x >= n) return;
  if (flags[x] == FLAG) return;
  flags[x] = FLAG;
  if (pos[x] < pos[x - 1]) add(x, d);
}
inline void init() {
  memset(flags, 0, sizeof flags);
  memset(f, 0, sizeof f);
  FLAG = 666;
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &a[i]);
    --a[i];
    pos[a[i]] = i;
  }
  for (int i = 1; i < n; ++i) {
    check(i, 1);
  }
}
int main() {
  init();
  scanf("%d", &m);
  for (int i = 0; i < m; ++i) {
    scanf("%d%d%d", &cmd, &l, &r);
    --l, --r;
    if (cmd == 2) {
      ++FLAG;
      check(a[l], -1);
      check(a[r], -1);
      check(a[l] + 1, -1);
      check(a[r] + 1, -1);
      swap(pos[a[l]], pos[a[r]]);
      swap(a[l], a[r]);
      ++FLAG;
      check(a[l], +1);
      check(a[r], +1);
      check(a[l] + 1, +1);
      check(a[r] + 1, +1);
      continue;
    }
    int res = get(l + 1, r) + 1;
    printf("%d\n", res);
  }
  return 0;
}
