#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:1000000000,1000000000")
using namespace std;
const long long inf = 1e18 + 7;
const long long mod = 1e9 + 7;
const double eps = 1e-12;
const double PI = 2 * acos(0.0);
const double E = 2.71828;
int n, q, sums[300005], where[300005], num[300005];
long long t[1200000], add[1200000];
long long get(long long v, long long tl, long long tr, long long l,
              long long r);
inline void calc() {
  for (long long i = 2; i <= n; i++) {
    sums[i] = sums[i - 1] + (where[i] < where[i - 1]);
  }
}
void build(long long v, long long l, long long r) {
  if (l == r)
    t[v] = sums[l];
  else {
    long long m = (l + r) >> 1LL;
    build(v + v, l, m);
    build(v + v + 1, m + 1, r);
    t[v] = t[v + v] + t[v + v + 1];
  }
}
void modify(long long v, long long tl, long long tr, long long l, long long r,
            long long val) {
  if (l <= tl && tr <= r) {
    t[v] += (tr - tl + 1) * val;
    add[v] += val;
    return;
  }
  long long tm = (tl + tr) >> 1LL;
  if (l <= tm) modify(v + v, tl, tm, l, r, val);
  if (r > tm) modify(v + v + 1, tm + 1, tr, l, r, val);
  t[v] = t[v + v] + t[v + v + 1] + (tr - tl + 1) * add[v];
}
long long get(long long v, long long tl, long long tr, long long l,
              long long r) {
  if (l <= tl && tr <= r) {
    return t[v];
  }
  long long tm = (tl + tr) >> 1LL;
  long long res = (min(tr, r) - max(tl, l) + 1) * add[v];
  if (l <= tm) res += get(v + v, tl, tm, l, r);
  if (r > tm) res += get(v + v + 1, tm + 1, tr, l, r);
  return res;
}
void change(long long num, long long prev, long long now,
            long long p_num = -1) {
  long long res_now = 0, res_next = 0;
  if (num - 1 != p_num && prev > where[num - 1] && now <= where[num - 1])
    res_now++;
  if (num - 1 != p_num && prev < where[num - 1] && now >= where[num - 1])
    res_now--;
  if (num + 1 != p_num && prev < where[num + 1] && now >= where[num + 1])
    res_next++;
  if (num + 1 != p_num && prev > where[num + 1] && now <= where[num + 1])
    res_next--;
  modify(1, 0, n + 1, num, n, res_now);
  modify(1, 0, n + 1, num + 1, n, res_next);
}
int main(void) {
  scanf("%d", &n);
  for (long long(i) = 0; (i) < (long long)(n); (i)++) {
    int x;
    scanf("%d", &x);
    where[x] = i + 1;
    num[i + 1] = x;
  }
  where[n + 1] = n + 1;
  num[n + 1] = n + 1;
  calc();
  build(1, 0, n + 1);
  scanf("%d", &q);
  for (long long(i) = 0; (i) < (long long)(q); (i)++) {
    int type, x, y;
    scanf("%d%d%d", &type, &x, &y);
    if (type == 1) {
      printf("%d\n", get(1, 0, n + 1, y, y) - get(1, 0, n + 1, x, x) + 1);
    } else {
      long long prev1 = where[num[x]], now1 = y;
      long long prev2 = where[num[y]], now2 = x;
      change(num[x], prev1, now1);
      change(num[y], prev2, now2, num[x]);
      where[num[x]] = y;
      where[num[y]] = x;
      swap(num[x], num[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>
#pragma comment(linker, "/STACK:1000000000,1000000000")
using namespace std;
const long long inf = 1e18 + 7;
const long long mod = 1e9 + 7;
const double eps = 1e-12;
const double PI = 2 * acos(0.0);
const double E = 2.71828;
int n, q, sums[300005], where[300005], num[300005];
long long t[1200000], add[1200000];
long long get(long long v, long long tl, long long tr, long long l,
              long long r);
inline void calc() {
  for (long long i = 2; i <= n; i++) {
    sums[i] = sums[i - 1] + (where[i] < where[i - 1]);
  }
}
void build(long long v, long long l, long long r) {
  if (l == r)
    t[v] = sums[l];
  else {
    long long m = (l + r) >> 1LL;
    build(v + v, l, m);
    build(v + v + 1, m + 1, r);
    t[v] = t[v + v] + t[v + v + 1];
  }
}
void modify(long long v, long long tl, long long tr, long long l, long long r,
            long long val) {
  if (l <= tl && tr <= r) {
    t[v] += (tr - tl + 1) * val;
    add[v] += val;
    return;
  }
  long long tm = (tl + tr) >> 1LL;
  if (l <= tm) modify(v + v, tl, tm, l, r, val);
  if (r > tm) modify(v + v + 1, tm + 1, tr, l, r, val);
  t[v] = t[v + v] + t[v + v + 1] + (tr - tl + 1) * add[v];
}
long long get(long long v, long long tl, long long tr, long long l,
              long long r) {
  if (l <= tl && tr <= r) {
    return t[v];
  }
  long long tm = (tl + tr) >> 1LL;
  long long res = (min(tr, r) - max(tl, l) + 1) * add[v];
  if (l <= tm) res += get(v + v, tl, tm, l, r);
  if (r > tm) res += get(v + v + 1, tm + 1, tr, l, r);
  return res;
}
void change(long long num, long long prev, long long now,
            long long p_num = -1) {
  long long res_now = 0, res_next = 0;
  if (num - 1 != p_num && prev > where[num - 1] && now <= where[num - 1])
    res_now++;
  if (num - 1 != p_num && prev < where[num - 1] && now >= where[num - 1])
    res_now--;
  if (num + 1 != p_num && prev < where[num + 1] && now >= where[num + 1])
    res_next++;
  if (num + 1 != p_num && prev > where[num + 1] && now <= where[num + 1])
    res_next--;
  modify(1, 0, n + 1, num, n, res_now);
  modify(1, 0, n + 1, num + 1, n, res_next);
}
int main(void) {
  scanf("%d", &n);
  for (long long(i) = 0; (i) < (long long)(n); (i)++) {
    int x;
    scanf("%d", &x);
    where[x] = i + 1;
    num[i + 1] = x;
  }
  where[n + 1] = n + 1;
  num[n + 1] = n + 1;
  calc();
  build(1, 0, n + 1);
  scanf("%d", &q);
  for (long long(i) = 0; (i) < (long long)(q); (i)++) {
    int type, x, y;
    scanf("%d%d%d", &type, &x, &y);
    if (type == 1) {
      printf("%d\n", get(1, 0, n + 1, y, y) - get(1, 0, n + 1, x, x) + 1);
    } else {
      long long prev1 = where[num[x]], now1 = y;
      long long prev2 = where[num[y]], now2 = x;
      change(num[x], prev1, now1);
      change(num[y], prev2, now2, num[x]);
      where[num[x]] = y;
      where[num[y]] = x;
      swap(num[x], num[y]);
    }
  }
  return 0;
}

```