#include <bits/stdc++.h>
using namespace std;
const int inf = 1e9;
const double eps = 1e-9;
const double INF = inf;
const double EPS = eps;
int P[310000], R[310000];
int A[(1 << 20) + 100], n;
void add(int v, int d) {
  v += (1 << 19);
  while (v) A[v] += d, v >>= 1;
}
int sum(int l, int r) {
  l += (1 << 19), r += (1 << 19);
  int ret = 0;
  while (l < r) {
    if (l & 1) ret += A[l], l++;
    if (r & 1) r--, ret += A[r];
    l >>= 1, r >>= 1;
  }
  return ret;
}
void recalc(int v, int w) {
  if (v > 0 && P[v - 1] > P[v]) add(v - 1, w);
  if (v < n - 1 && P[v] > P[v + 1]) add(v, w);
}
int main() {
  int i, q, x, y, tp;
  scanf("%d", &n);
  for (i = 0; i < n; i++) scanf("%d", &R[i]), R[i]--, P[R[i]] = i;
  for (i = 0; i < n - 1; i++)
    if (P[i] > P[i + 1]) add(i, 1);
  scanf("%d", &q);
  while (q > 0) {
    q--, scanf("%d%d%d", &tp, &x, &y), x--, y--;
    if (tp == 1) {
      printf("%d\n", sum(x, y) + 1);
      continue;
    }
    x = R[x], y = R[y];
    if (x > y) swap(x, y);
    recalc(x, -1);
    recalc(y, -1);
    if (x + 1 == y && P[x] > P[y]) add(x, 1);
    swap(P[x], P[y]), R[P[x]] = x, R[P[y]] = y;
    recalc(x, 1);
    recalc(y, 1);
    if (x > y) swap(x, y);
    if (x + 1 == y && P[x] > P[y]) add(x, -1);
  };
  return 0;
}

### Prompt

Create a solution in CPP for 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 = 1e9;
const double eps = 1e-9;
const double INF = inf;
const double EPS = eps;
int P[310000], R[310000];
int A[(1 << 20) + 100], n;
void add(int v, int d) {
  v += (1 << 19);
  while (v) A[v] += d, v >>= 1;
}
int sum(int l, int r) {
  l += (1 << 19), r += (1 << 19);
  int ret = 0;
  while (l < r) {
    if (l & 1) ret += A[l], l++;
    if (r & 1) r--, ret += A[r];
    l >>= 1, r >>= 1;
  }
  return ret;
}
void recalc(int v, int w) {
  if (v > 0 && P[v - 1] > P[v]) add(v - 1, w);
  if (v < n - 1 && P[v] > P[v + 1]) add(v, w);
}
int main() {
  int i, q, x, y, tp;
  scanf("%d", &n);
  for (i = 0; i < n; i++) scanf("%d", &R[i]), R[i]--, P[R[i]] = i;
  for (i = 0; i < n - 1; i++)
    if (P[i] > P[i + 1]) add(i, 1);
  scanf("%d", &q);
  while (q > 0) {
    q--, scanf("%d%d%d", &tp, &x, &y), x--, y--;
    if (tp == 1) {
      printf("%d\n", sum(x, y) + 1);
      continue;
    }
    x = R[x], y = R[y];
    if (x > y) swap(x, y);
    recalc(x, -1);
    recalc(y, -1);
    if (x + 1 == y && P[x] > P[y]) add(x, 1);
    swap(P[x], P[y]), R[P[x]] = x, R[P[y]] = y;
    recalc(x, 1);
    recalc(y, 1);
    if (x > y) swap(x, y);
    if (x + 1 == y && P[x] > P[y]) add(x, -1);
  };
  return 0;
}

```