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;
template <class T>
inline void read(T& x) {
  bool fu = 0;
  char c;
  for (c = getchar(); c <= 32; c = getchar())
    ;
  if (c == '-') fu = 1, c = getchar();
  for (x = 0; c > 32; c = getchar()) x = x * 10 + c - '0';
  if (fu) x = -x;
};
template <class T>
inline void read(T& x, T& y) {
  read(x);
  read(y);
}
template <class T>
inline void read(T& x, T& y, T& z) {
  read(x);
  read(y);
  read(z);
}
inline char getc() {
  char c;
  for (c = getchar(); c <= 32; c = getchar())
    ;
  return c;
}
const int N = 300010;
int n, m, i, j, k, l, p, x, y;
int a[N], pos[N];
int s[N];
void A(int i, int t) {
  for (; i <= n; i += i & -i) s[i] += t;
}
int S(int i) {
  int ans = 0;
  for (; i; i -= i & -i) ans += s[i];
  return ans;
}
void h(int x, int t) {
  if (pos[x] - pos[x - 1] < 0) A(x, -1);
  if (pos[x + 1] - pos[x] < 0) A(x + 1, -1);
  pos[x] = t;
  if (pos[x] - pos[x - 1] < 0) A(x, 1);
  if (pos[x + 1] - pos[x] < 0) A(x + 1, 1);
}
int main() {
  read(n);
  pos[0] = pos[n + 1] = n + 1;
  for (i = 1; i <= n; i++) read(a[i]), pos[a[i]] = i;
  for (i = 1; i <= n; i++)
    if (pos[i] - pos[i - 1] < 0) {
      A(i, 1);
    }
  for (read(m); m; m--) {
    int t;
    read(t);
    if (t == 1) {
      read(x, y);
      printf("%d\n", S(y) - S(x) + 1);
    } else if (t == 2) {
      read(x, y);
      h(a[x], y);
      h(a[y], x);
      swap(a[x], a[y]);
    }
  }
  return 0;
}
