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 N = 300000 + 10;
struct que {
  int a, l, r;
} q[N];
int a[N], whe[N];
int ans[N];
int dp[N], n;
int c[N];
void add(int u, int val) {
  while (u < N) {
    c[u] += val;
    u += ((u) & (-u));
  }
}
int ask(int u) {
  int ret = 0;
  while (u) {
    ret += c[u];
    u -= ((u) & (-u));
  }
  return ret;
}
void fix(int u) {
  if (u == 1) return;
  if (u > n) return;
  int now;
  int last = ask(u) - ask(u - 1);
  if (whe[u] < whe[u - 1])
    now = 1;
  else if (whe[u] > whe[u - 1])
    now = 0;
  add(u, now - last);
}
void solve(int l, int r) {
  if (l == r) {
    if (q[l].a == 1)
      ans[l] = ask(q[l].r) - ask(q[l].l) + 1;
    else {
      swap(whe[a[q[l].l]], whe[a[q[l].r]]);
      swap(a[q[l].l], a[q[l].r]);
      fix(a[q[l].l]);
      fix(a[q[l].l] + 1);
      fix(a[q[r].r]);
      fix(a[q[r].r] + 1);
    }
    return;
  }
  int mid = (l + r) >> 1;
  solve(l, mid);
  if (mid + 1 <= r) solve(mid + 1, r);
}
int main() {
  int i;
  scanf("%d", &n);
  for (i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
    whe[a[i]] = i;
  }
  memset(c, 0, sizeof(c));
  for (i = 2; i <= n; ++i)
    if (whe[i - 1] > whe[i]) add(i, 1);
  int Q;
  scanf("%d", &Q);
  for (i = 1; i <= Q; ++i) scanf("%d%d%d", &q[i].a, &q[i].l, &q[i].r);
  solve(1, Q);
  for (i = 1; i <= Q; ++i)
    if (q[i].a == 1) printf("%d\n", ans[i]);
  return 0;
}
