#include <bits/stdc++.h>
using namespace std;
const int N = 300006;
int n, A[N], q, pos[N];
struct Interval_Tree {
  struct data {
    int s, t, Lval, Rval, sum;
    void init(int a, int b, int c, int d, int e) {
      s = a, t = b, Lval = c, Rval = d, sum = e;
    }
  } T[N << 2];
  inline void pushUp(data &fa, data &L, data &R) {
    fa.s = L.s, fa.t = R.t;
    fa.Lval = L.Lval, fa.Rval = R.Rval;
    fa.sum = L.sum + R.sum;
    if (L.Rval <= R.Lval) fa.sum--;
  }
  void build(int id, int L, int R) {
    T[id].init(L, R, pos[L], pos[R], 1);
    if (L == R) return;
    int mid = (L + R) >> 1;
    build(id << 1, L, mid), build(id << 1 | 1, mid + 1, R);
    pushUp(T[id], T[id << 1], T[id << 1 | 1]);
  }
  void update(int id, int p, int val) {
    if (T[id].s == T[id].t) {
      T[id].Lval = T[id].Rval = val;
      return;
    }
    int mid = (T[id].s + T[id].t) >> 1;
    if (p <= mid)
      update(id << 1, p, val);
    else
      update(id << 1 | 1, p, val);
    pushUp(T[id], T[id << 1], T[id << 1 | 1]);
  }
  data query(int id, int L, int R) {
    if (L <= T[id].s && T[id].t <= R) return T[id];
    int mid = (T[id].s + T[id].t) >> 1;
    if (R <= mid)
      return query(id << 1, L, R);
    else if (L > mid)
      return query(id << 1 | 1, L, R);
    data f1 = query(id << 1, L, R);
    data f2 = query(id << 1 | 1, L, R);
    data f3;
    pushUp(f3, f1, f2);
    return f3;
  }
} tree;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &A[i]), pos[A[i]] = i;
  tree.build(1, 1, n);
  scanf("%d", &q);
  for (int i = 0, fg, x, y; i < q; i++) {
    scanf("%d%d%d", &fg, &x, &y);
    if (fg == 1)
      printf("%d\n", tree.query(1, x, y).sum);
    else {
      tree.update(1, A[x], y), tree.update(1, A[y], x);
      swap(A[x], A[y]);
    }
  }
  return 0;
}

### 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 N = 300006;
int n, A[N], q, pos[N];
struct Interval_Tree {
  struct data {
    int s, t, Lval, Rval, sum;
    void init(int a, int b, int c, int d, int e) {
      s = a, t = b, Lval = c, Rval = d, sum = e;
    }
  } T[N << 2];
  inline void pushUp(data &fa, data &L, data &R) {
    fa.s = L.s, fa.t = R.t;
    fa.Lval = L.Lval, fa.Rval = R.Rval;
    fa.sum = L.sum + R.sum;
    if (L.Rval <= R.Lval) fa.sum--;
  }
  void build(int id, int L, int R) {
    T[id].init(L, R, pos[L], pos[R], 1);
    if (L == R) return;
    int mid = (L + R) >> 1;
    build(id << 1, L, mid), build(id << 1 | 1, mid + 1, R);
    pushUp(T[id], T[id << 1], T[id << 1 | 1]);
  }
  void update(int id, int p, int val) {
    if (T[id].s == T[id].t) {
      T[id].Lval = T[id].Rval = val;
      return;
    }
    int mid = (T[id].s + T[id].t) >> 1;
    if (p <= mid)
      update(id << 1, p, val);
    else
      update(id << 1 | 1, p, val);
    pushUp(T[id], T[id << 1], T[id << 1 | 1]);
  }
  data query(int id, int L, int R) {
    if (L <= T[id].s && T[id].t <= R) return T[id];
    int mid = (T[id].s + T[id].t) >> 1;
    if (R <= mid)
      return query(id << 1, L, R);
    else if (L > mid)
      return query(id << 1 | 1, L, R);
    data f1 = query(id << 1, L, R);
    data f2 = query(id << 1 | 1, L, R);
    data f3;
    pushUp(f3, f1, f2);
    return f3;
  }
} tree;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &A[i]), pos[A[i]] = i;
  tree.build(1, 1, n);
  scanf("%d", &q);
  for (int i = 0, fg, x, y; i < q; i++) {
    scanf("%d%d%d", &fg, &x, &y);
    if (fg == 1)
      printf("%d\n", tree.query(1, x, y).sum);
    else {
      tree.update(1, A[x], y), tree.update(1, A[y], x);
      swap(A[x], A[y]);
    }
  }
  return 0;
}

```