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 long long maxn = 300000 + 100;
long long n, p[maxn], flag[maxn];
long long b[maxn], a[maxn];
void init() {
  for (long long i = 1; i <= n; i++) {
    b[i] = 0;
    flag[i] = 0;
  }
}
long long lowbit(long long x) { return x & (-x); }
void add(long long x, long long v) {
  for (long long i = x; i <= n; i += lowbit(i)) {
    b[i] += v;
  }
}
long long ask(long long x) {
  long long re = 0;
  for (long long i = x; i >= 1; i -= lowbit(i)) {
    re += b[i];
  }
  return re;
}
void solve(long long x, long long y, long long st1, long long st2) {
  if (st1) {
    if (p[x] > p[x - 1]) {
      if (!flag[x]) add(x, -1);
    }
  } else {
    if (p[x] < p[x - 1]) {
      if (!flag[x]) add(x, 1);
    }
  }
  if (st2) {
    if (p[x + 1] > p[x]) {
      if (!flag[x + 1]) add(x + 1, -1);
    }
  } else {
    if (p[x + 1] < p[x]) {
      if (!flag[x + 1]) add(x + 1, 1);
    }
  }
}
int main() {
  scanf("%lld", &n);
  init();
  a[0] = 1 << 30;
  p[0] = 1 << 30;
  for (long long i = 1; i <= n; i++) {
    scanf("%lld", &a[i]);
    p[a[i]] = i;
  }
  for (long long i = 1; i <= n; i++) {
    if (p[i] < p[i - 1]) add(i, 1);
  }
  long long k, q, x, y;
  scanf("%lld", &k);
  for (long long i = 0; i < k; i++) {
    scanf("%lld%lld%lld", &q, &x, &y);
    if (q == 1) {
      printf("%lld\n", ask(y) - ask(x) + 1);
    } else {
      long long vx = a[x], vy = a[y];
      long long st1 = p[vx] < p[vx - 1] ? 1 : 0;
      long long st2 = p[vx + 1] < p[vx] ? 1 : 0;
      long long st3 = p[vy] < p[vy - 1] ? 1 : 0;
      long long st4 = p[vy + 1] < p[vy] ? 1 : 0;
      p[a[x]] = y;
      p[a[y]] = x;
      swap(a[x], a[y]);
      solve(vx, vy, st1, st2);
      flag[vx] = flag[vx + 1] = 1;
      solve(vy, vx, st3, st4);
      flag[vx] = flag[vx + 1] = 0;
    }
  }
  return 0;
}
