#include <bits/stdc++.h>
using namespace std;
const int MaxN = 300005;
const double eps = 1e-7;
const double DINF = 1e100;
const int INF = 1000000006;
const long long LINF = 1000000000000000005ll;
int n;
int a[MaxN], p[MaxN];
struct Tnode {
  int vl, vr, res;
} T[MaxN * 4];
void update(Tnode &A, Tnode B, Tnode C) {
  A.vl = B.vl;
  A.vr = C.vr;
  A.res = B.res + C.res + (B.vr > C.vl);
}
void build(int p, int l, int r) {
  if (l == r) {
    T[p].vl = T[p].vr = a[r];
    return;
  }
  int mid = (l + r) >> 1;
  build(((p << 1)), l, mid);
  build(((p << 1) + 1), mid + 1, r);
  update(T[p], T[((p << 1))], T[((p << 1) + 1)]);
}
void modify(int p, int l, int r, int pos, int v) {
  if (l == r) {
    T[p].vl = T[p].vr = v;
    return;
  }
  int mid = (l + r) >> 1;
  if (pos <= mid)
    modify(((p << 1)), l, mid, pos, v);
  else
    modify(((p << 1) + 1), mid + 1, r, pos, v);
  update(T[p], T[((p << 1))], T[((p << 1) + 1)]);
}
Tnode query(int p, int l, int r, int i, int j) {
  if (i == l && j == r) return T[p];
  int mid = (l + r) >> 1;
  if (j <= mid)
    return query(((p << 1)), l, mid, i, j);
  else if (i > mid)
    return query(((p << 1) + 1), mid + 1, r, i, j);
  else {
    Tnode res;
    update(res, query(((p << 1)), l, mid, i, mid),
           query(((p << 1) + 1), mid + 1, r, mid + 1, j));
    return res;
  }
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    scanf("%d", p + i);
    a[p[i]] = i;
  }
  build(1, 1, n);
  int Q;
  cin >> Q;
  while (Q--) {
    int p, x, y;
    scanf("%d%d%d", &p, &x, &y);
    if (p == 1) {
      printf("%d\n", query(1, 1, n, x, y).res + 1);
    } else {
      modify(1, 1, n, ::p[x], y);
      modify(1, 1, n, ::p[y], x);
      swap(::p[x], ::p[y]);
      a[::p[x]] = x;
      a[::p[y]] = y;
    }
  }
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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 MaxN = 300005;
const double eps = 1e-7;
const double DINF = 1e100;
const int INF = 1000000006;
const long long LINF = 1000000000000000005ll;
int n;
int a[MaxN], p[MaxN];
struct Tnode {
  int vl, vr, res;
} T[MaxN * 4];
void update(Tnode &A, Tnode B, Tnode C) {
  A.vl = B.vl;
  A.vr = C.vr;
  A.res = B.res + C.res + (B.vr > C.vl);
}
void build(int p, int l, int r) {
  if (l == r) {
    T[p].vl = T[p].vr = a[r];
    return;
  }
  int mid = (l + r) >> 1;
  build(((p << 1)), l, mid);
  build(((p << 1) + 1), mid + 1, r);
  update(T[p], T[((p << 1))], T[((p << 1) + 1)]);
}
void modify(int p, int l, int r, int pos, int v) {
  if (l == r) {
    T[p].vl = T[p].vr = v;
    return;
  }
  int mid = (l + r) >> 1;
  if (pos <= mid)
    modify(((p << 1)), l, mid, pos, v);
  else
    modify(((p << 1) + 1), mid + 1, r, pos, v);
  update(T[p], T[((p << 1))], T[((p << 1) + 1)]);
}
Tnode query(int p, int l, int r, int i, int j) {
  if (i == l && j == r) return T[p];
  int mid = (l + r) >> 1;
  if (j <= mid)
    return query(((p << 1)), l, mid, i, j);
  else if (i > mid)
    return query(((p << 1) + 1), mid + 1, r, i, j);
  else {
    Tnode res;
    update(res, query(((p << 1)), l, mid, i, mid),
           query(((p << 1) + 1), mid + 1, r, mid + 1, j));
    return res;
  }
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    scanf("%d", p + i);
    a[p[i]] = i;
  }
  build(1, 1, n);
  int Q;
  cin >> Q;
  while (Q--) {
    int p, x, y;
    scanf("%d%d%d", &p, &x, &y);
    if (p == 1) {
      printf("%d\n", query(1, 1, n, x, y).res + 1);
    } else {
      modify(1, 1, n, ::p[x], y);
      modify(1, 1, n, ::p[y], x);
      swap(::p[x], ::p[y]);
      a[::p[x]] = x;
      a[::p[y]] = y;
    }
  }
  return 0;
}

```