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 maxn = 3 * 100500;
const int infi = (1 << 30);
const long long infl = (1LL << 62);
const double eps = 1e-9;
const long long mod = 1000000007LL;
const unsigned long long hash_base = 257LL;
const double pi = acos(-1.0);
int n, q, ct, cx, cy;
int a[maxn], b[maxn];
int t[4 * maxn];
void update(int v, int tl, int tr, int pos, int val) {
  if (tl == tr)
    t[v] = val;
  else {
    int tm = (tl + tr) / 2;
    if (pos <= tm)
      update(v * 2 + 1, tl, tm, pos, val);
    else
      update(v * 2 + 2, tm + 1, tr, pos, val);
    t[v] = t[v * 2 + 1] + t[v * 2 + 2];
  }
}
int sum(int v, int tl, int tr, int l, int r) {
  if (l > r) return 0;
  if (tl == l && tr == r) return t[v];
  int tm = (tl + tr) / 2;
  return sum(v * 2 + 1, tl, tm, l, min(tm, r)) +
         sum(v * 2 + 2, tm + 1, tr, max(tm + 1, l), r);
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> b[i];
    b[i]--;
    a[b[i]] = i;
  }
  for (int i = 0; i < 4 * maxn; i++) t[i] = 0;
  for (int i = 0; i < n - 1; i++)
    if (a[i] > a[i + 1]) update(0, 0, n - 1, i, 1);
  cin >> q;
  while (q--) {
    cin >> ct >> cx >> cy;
    --cx;
    --cy;
    if (ct == 1) {
      cout << sum(0, 0, n - 1, cx, cy - 1) + 1 << endl;
    } else {
      swap(a[b[cx]], a[b[cy]]);
      if (b[cx] < n - 1) {
        if (a[b[cx]] > a[b[cx] + 1])
          update(0, 0, n - 1, b[cx], 1);
        else
          update(0, 0, n - 1, b[cx], 0);
      }
      if (b[cx] > 0) {
        if (a[b[cx] - 1] > a[b[cx]])
          update(0, 0, n - 1, b[cx] - 1, 1);
        else
          update(0, 0, n - 1, b[cx] - 1, 0);
      }
      swap(cx, cy);
      if (b[cx] < n - 1) {
        if (a[b[cx]] > a[b[cx] + 1])
          update(0, 0, n - 1, b[cx], 1);
        else
          update(0, 0, n - 1, b[cx], 0);
      }
      if (b[cx] > 0) {
        if (a[b[cx] - 1] > a[b[cx]])
          update(0, 0, n - 1, b[cx] - 1, 1);
        else
          update(0, 0, n - 1, b[cx] - 1, 0);
      }
      swap(b[cx], b[cy]);
    }
  }
  return 0;
}
