#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const double EPS = 1e-8;
const int MAXN = 3e5 + 54;
int sign(double x) { return x < -EPS ? -1 : x > EPS; }
int dbcmp(double l, double r) { return sign(l - r); }
int a[MAXN], pos[MAXN], tree[MAXN << 2];
void build(int rt, int l, int r) {
  if (l == r) {
    tree[rt] = pos[l] < pos[l - 1];
    return;
  }
  int mid = l + r >> 1;
  build(rt << 1, l, mid);
  build(rt << 1 | 1, mid + 1, r);
  tree[rt] = tree[rt << 1] + tree[rt << 1 | 1];
}
int query(int rt, int L, int R, int l, int r) {
  if (L > r || l > R) return 0;
  if (l <= L && R <= r) return tree[rt];
  int mid = L + R >> 1, ans = 0;
  if (l <= mid) ans += query(rt << 1, L, mid, l, r);
  if (r > mid) ans += query(rt << 1 | 1, mid + 1, R, l, r);
  return ans;
}
void update(int rt, int l, int r, int x, int v) {
  if (l == r) {
    tree[rt] = v;
    return;
  }
  int mid = l + r >> 1;
  if (x <= mid)
    update(rt << 1, l, mid, x, v);
  else
    update(rt << 1 | 1, mid + 1, r, x, v);
  tree[rt] = tree[rt << 1] + tree[rt << 1 | 1];
}
int main() {
  int n, q;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    pos[a[i]] = i;
  }
  scanf("%d", &q);
  build(1, 1, n + 1);
  for (int i = 1; i <= q; i++) {
    int op, x, y;
    scanf("%d%d%d", &op, &x, &y);
    if (op == 1)
      printf("%d\n", x == y ? 1 : (1 + query(1, 1, n + 1, x + 1, y)));
    else {
      int pos1 = a[x], pos2 = a[y];
      swap(a[x], a[y]);
      swap(pos[pos1], pos[pos2]);
      update(1, 1, n + 1, pos1, pos[pos1 - 1] > pos[pos1]);
      update(1, 1, n + 1, pos1 + 1, pos[pos1] > pos[pos1 + 1]);
      update(1, 1, n + 1, pos2 + 1, pos[pos2] > pos[pos2 + 1]);
      update(1, 1, n + 1, pos2, pos[pos2 - 1] > pos[pos2]);
    }
  }
}

### Prompt

Your challenge is to write a CPP solution to the following problem:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const double EPS = 1e-8;
const int MAXN = 3e5 + 54;
int sign(double x) { return x < -EPS ? -1 : x > EPS; }
int dbcmp(double l, double r) { return sign(l - r); }
int a[MAXN], pos[MAXN], tree[MAXN << 2];
void build(int rt, int l, int r) {
  if (l == r) {
    tree[rt] = pos[l] < pos[l - 1];
    return;
  }
  int mid = l + r >> 1;
  build(rt << 1, l, mid);
  build(rt << 1 | 1, mid + 1, r);
  tree[rt] = tree[rt << 1] + tree[rt << 1 | 1];
}
int query(int rt, int L, int R, int l, int r) {
  if (L > r || l > R) return 0;
  if (l <= L && R <= r) return tree[rt];
  int mid = L + R >> 1, ans = 0;
  if (l <= mid) ans += query(rt << 1, L, mid, l, r);
  if (r > mid) ans += query(rt << 1 | 1, mid + 1, R, l, r);
  return ans;
}
void update(int rt, int l, int r, int x, int v) {
  if (l == r) {
    tree[rt] = v;
    return;
  }
  int mid = l + r >> 1;
  if (x <= mid)
    update(rt << 1, l, mid, x, v);
  else
    update(rt << 1 | 1, mid + 1, r, x, v);
  tree[rt] = tree[rt << 1] + tree[rt << 1 | 1];
}
int main() {
  int n, q;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    pos[a[i]] = i;
  }
  scanf("%d", &q);
  build(1, 1, n + 1);
  for (int i = 1; i <= q; i++) {
    int op, x, y;
    scanf("%d%d%d", &op, &x, &y);
    if (op == 1)
      printf("%d\n", x == y ? 1 : (1 + query(1, 1, n + 1, x + 1, y)));
    else {
      int pos1 = a[x], pos2 = a[y];
      swap(a[x], a[y]);
      swap(pos[pos1], pos[pos2]);
      update(1, 1, n + 1, pos1, pos[pos1 - 1] > pos[pos1]);
      update(1, 1, n + 1, pos1 + 1, pos[pos1] > pos[pos1 + 1]);
      update(1, 1, n + 1, pos2 + 1, pos[pos2] > pos[pos2 + 1]);
      update(1, 1, n + 1, pos2, pos[pos2 - 1] > pos[pos2]);
    }
  }
}

```