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
#include <bits/stdc++.h>
#pragma warning(disable : 4996)
#pragma comment(linker, "/STACK:336777216")
using namespace std;
const unsigned MAX = 1e5 + 5;
const long long MOD = 1e9 + 7;
const int INF = 0x3f3f3f3f;
const long long LLINF = 0x3f3f3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int mov[][3] = {{0, 0},  {1, 0},  {1, 1},  {1, -1}, {0, 1},
                      {0, -1}, {-1, 1}, {-1, 0}, {-1, -1}};
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
inline long long g() {
  char c;
  long long p = 1;
  while (!isdigit(c = getchar()))
    if (c == '-') p = -p;
  long long x = c - '0';
  while (isdigit(c = getchar())) x = x * 10 + c - '0';
  return p * x;
}
int n;
int Vi[300005];
int pos[300005];
int cnt[300005];
int vis[300005];
void update(int i, int delta) {
  while (i <= n) {
    cnt[i] += delta;
    i += i & (-i);
  }
}
int getsum(int i) {
  int sum = 0;
  while (i > 0) {
    sum += cnt[i];
    i -= i & (-i);
  }
  return sum;
}
int query(int l, int r) { return getsum(r) - getsum(l - 1); }
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  bool flag = 0;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> Vi[i];
    pos[Vi[i]] = i;
  }
  pos[0] = INF;
  for (int i = 1; i <= n; i++) {
    if (pos[i - 1] > pos[i]) {
      update(i, 1);
      vis[i] = 1;
    }
  }
  int q;
  cin >> q;
  while (q--) {
    int ty;
    cin >> ty;
    if (ty == 2) {
      int X, Y;
      cin >> X >> Y;
      pos[Vi[X]] = Y;
      pos[Vi[Y]] = X;
      swap(Vi[X], Vi[Y]);
      {
        int u = Vi[Y];
        if (u > 1) {
          if (pos[u - 1] < pos[u]) {
            update(u, 0 - vis[u]);
            vis[u] = 0;
          } else {
            update(u, 1 - vis[u]);
            vis[u] = 1;
          }
        }
        if (u < n) {
          if (pos[u] < pos[u + 1]) {
            update(u + 1, 0 - vis[u + 1]);
            vis[u + 1] = 0;
          } else {
            update(u + 1, 1 - vis[u + 1]);
            vis[u + 1] = 1;
          }
        }
      }
      {
        int u = Vi[X];
        if (u > 1) {
          if (pos[u - 1] < pos[u]) {
            update(u, 0 - vis[u]);
            vis[u] = 0;
          } else {
            update(u, 1 - vis[u]);
            vis[u] = 1;
          }
        }
        if (u < n) {
          if (pos[u] < pos[u + 1]) {
            update(u + 1, 0 - vis[u + 1]);
            vis[u + 1] = 0;
          } else {
            update(u + 1, 1 - vis[u + 1]);
            vis[u + 1] = 1;
          }
        }
      }
    } else {
      int l, r;
      cin >> l >> r;
      cout << query(l + 1, r) + 1 << '\n';
    }
  }
}
