#include <bits/stdc++.h>
using namespace std;
inline int in() {
  int32_t x;
  scanf("%d", &x);
  return x;
}
inline long long lin() {
  long long x;
  scanf("%lld", &x);
  return x;
}
inline string get() {
  char ch[2000010];
  scanf("%s", ch);
  return ch;
}
inline void read(int *a, int n) {
  for (int i = 0; i < n; i++) a[i] = in();
}
template <class blank>
inline void out(blank x) {
  cout << x << "\n";
  exit(0);
}
const int maxn = 3e5 + 10;
const int maxm = 1e6 + 10;
const int maxlg = 20;
const int base = 29;
const int base2 = 31;
const int mod2 = 1e9 + 9;
const int mod = 1e9 + 7;
const long long INF = 1e18 + 10;
const int SQ = 317 + 5;
int a[maxn], n;
int stree[4 * maxn];
int where[maxn];
inline void build(int p = 0, int l = 0, int r = n - 2) {
  if (l == r) {
    stree[p] = where[l + 1] < where[l];
    return;
  }
  int mid = (l + r) >> 1;
  build(2 * p + 1, l, mid);
  build(2 * p + 2, mid + 1, r);
  stree[p] = stree[2 * p + 1] + stree[2 * p + 2];
}
inline int get(int ql, int qr, int p = 0, int l = 0, int r = n - 2) {
  if (l > qr || r < ql) return 0;
  if (l >= ql && r <= qr) return stree[p];
  int mid = (l + r) >> 1;
  return get(ql, qr, 2 * p + 1, l, mid) + get(ql, qr, 2 * p + 2, mid + 1, r);
}
inline void update(int qx, int p = 0, int l = 0, int r = n - 2) {
  if (l > qx || r < qx) return;
  if (l == r) {
    stree[p] = where[l + 1] < where[l];
    return;
  }
  int mid = (l + r) >> 1;
  update(qx, 2 * p + 1, l, mid), update(qx, 2 * p + 2, mid + 1, r);
  stree[p] = stree[2 * p + 1] + stree[2 * p + 2];
}
int32_t main() {
  n = in();
  memset(where, -1, sizeof where);
  read(a, n);
  for (int i = 0; i < n; i++) where[--a[i]] = i;
  build();
  int q = in();
  while (q--) {
    int tp = in(), a = in() - 1, b = in() - 1;
    if (tp == 1) {
      b--;
      cout << get(a, b) + 1 << "\n";
      continue;
    }
    swap(::a[a], ::a[b]);
    swap(where[::a[a]], where[::a[b]]);
    int q[4] = {::a[a], ::a[b], ::a[a] - 1, ::a[b] - 1};
    for (int pt = 0; pt < 4; pt++) {
      if (q[pt] >= 0 && q[pt] < n - 1) {
        update(q[pt]);
      }
    }
  }
}

### Prompt

Please create a solution in cpp 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;
inline int in() {
  int32_t x;
  scanf("%d", &x);
  return x;
}
inline long long lin() {
  long long x;
  scanf("%lld", &x);
  return x;
}
inline string get() {
  char ch[2000010];
  scanf("%s", ch);
  return ch;
}
inline void read(int *a, int n) {
  for (int i = 0; i < n; i++) a[i] = in();
}
template <class blank>
inline void out(blank x) {
  cout << x << "\n";
  exit(0);
}
const int maxn = 3e5 + 10;
const int maxm = 1e6 + 10;
const int maxlg = 20;
const int base = 29;
const int base2 = 31;
const int mod2 = 1e9 + 9;
const int mod = 1e9 + 7;
const long long INF = 1e18 + 10;
const int SQ = 317 + 5;
int a[maxn], n;
int stree[4 * maxn];
int where[maxn];
inline void build(int p = 0, int l = 0, int r = n - 2) {
  if (l == r) {
    stree[p] = where[l + 1] < where[l];
    return;
  }
  int mid = (l + r) >> 1;
  build(2 * p + 1, l, mid);
  build(2 * p + 2, mid + 1, r);
  stree[p] = stree[2 * p + 1] + stree[2 * p + 2];
}
inline int get(int ql, int qr, int p = 0, int l = 0, int r = n - 2) {
  if (l > qr || r < ql) return 0;
  if (l >= ql && r <= qr) return stree[p];
  int mid = (l + r) >> 1;
  return get(ql, qr, 2 * p + 1, l, mid) + get(ql, qr, 2 * p + 2, mid + 1, r);
}
inline void update(int qx, int p = 0, int l = 0, int r = n - 2) {
  if (l > qx || r < qx) return;
  if (l == r) {
    stree[p] = where[l + 1] < where[l];
    return;
  }
  int mid = (l + r) >> 1;
  update(qx, 2 * p + 1, l, mid), update(qx, 2 * p + 2, mid + 1, r);
  stree[p] = stree[2 * p + 1] + stree[2 * p + 2];
}
int32_t main() {
  n = in();
  memset(where, -1, sizeof where);
  read(a, n);
  for (int i = 0; i < n; i++) where[--a[i]] = i;
  build();
  int q = in();
  while (q--) {
    int tp = in(), a = in() - 1, b = in() - 1;
    if (tp == 1) {
      b--;
      cout << get(a, b) + 1 << "\n";
      continue;
    }
    swap(::a[a], ::a[b]);
    swap(where[::a[a]], where[::a[b]]);
    int q[4] = {::a[a], ::a[b], ::a[a] - 1, ::a[b] - 1};
    for (int pt = 0; pt < 4; pt++) {
      if (q[pt] >= 0 && q[pt] < n - 1) {
        update(q[pt]);
      }
    }
  }
}

```