Problem: Serega loves fun. However, everyone has fun in the unique manner. Serega has fun by solving query problems. One day Fedor came up with such a problem.

You are given an array a consisting of n positive integers and queries to it. The queries can be of two types:

  1. Make a unit cyclic shift to the right on the segment from l to r (both borders inclusive). That is rearrange elements of the array in the following manner:a[l], a[l + 1], ..., a[r - 1], a[r] → a[r], a[l], a[l + 1], ..., a[r - 1].
  2. Count how many numbers equal to k are on the segment from l to r (both borders inclusive). 



Fedor hurried to see Serega enjoy the problem and Serega solved it really quickly. Let's see, can you solve it?

Input

The first line contains integer n (1 ≤ n ≤ 105) — the number of elements of the array. The second line contains n integers a[1], a[2], ..., a[n] (1 ≤ a[i] ≤ n).

The third line contains a single integer q (1 ≤ q ≤ 105) — the number of queries. The next q lines contain the queries.

As you need to respond to the queries online, the queries will be encoded. A query of the first type will be given in format: 1 l'i r'i. A query of the second type will be given in format: 2 l'i r'i k'i. All the number in input are integer. They satisfy the constraints: 1 ≤ l'i, r'i, k'i ≤ n.

To decode the queries from the data given in input, you need to perform the following transformations:

li = ((l'i + lastans - 1) mod n) + 1; ri = ((r'i + lastans - 1) mod n) + 1; ki = ((k'i + lastans - 1) mod n) + 1.

Where lastans is the last reply to the query of the 2-nd type (initially, lastans = 0). If after transformation li is greater than ri, you must swap these values.

Output

For each query of the 2-nd type print the answer on a single line.

Examples

Input

7
6 6 2 7 4 2 5
7
1 3 6
2 2 4 2
2 2 4 7
2 2 2 5
1 2 6
1 1 4
2 1 7 3


Output

2
1
0
0


Input

8
8 4 2 2 7 7 8 8
8
1 8 8
2 8 1 7
1 8 1
1 7 3
2 8 8 3
1 1 4
1 2 7
1 4 5


Output

2
0

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 10;
const int maxm = 320 + 10;
const int SQRT = 320;
int a[maxn], b[maxn][maxm];
deque<int> d[maxm];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n; i++) {
    d[i / SQRT].push_back(a[i]);
    b[a[i]][i / SQRT]++;
  }
  int m, ans = 0;
  cin >> m;
  while (m--) {
    int type, l, r;
    cin >> type >> l >> r;
    l--, r--;
    l = (l + ans) % n;
    r = (r + ans) % n;
    if (r < l) swap(l, r);
    if (type == 1) {
      int st = (l / SQRT);
      int fi = (r / SQRT);
      int shift = d[r / SQRT][r % SQRT];
      d[r / SQRT].erase(d[r / SQRT].begin() + r % SQRT);
      d[l / SQRT].insert(d[l / SQRT].begin() + l % SQRT, shift);
      b[shift][r / SQRT]--;
      b[shift][l / SQRT]++;
      for (int i = st; i < fi; i++) {
        d[i + 1].push_front(d[i].back());
        b[d[i].back()][i + 1]++;
        b[d[i].back()][i]--;
        d[i].pop_back();
      }
    } else {
      int k;
      cin >> k;
      k = (k + ans - 1) % n + 1;
      int st = (l / SQRT) + 1;
      int fi = (r / SQRT) - 1;
      ans = 0;
      for (int i = st; i <= fi; i++) ans += b[k][i];
      if (l / SQRT == r / SQRT) {
        for (int i = l % SQRT; i <= r % SQRT; i++) ans += (d[l / SQRT][i] == k);
      } else {
        for (int i = l % SQRT; i < d[l / SQRT].size(); i++)
          ans += (d[l / SQRT][i] == k);
        for (int i = 0; i <= r % SQRT; i++) ans += (d[r / SQRT][i] == k);
      }
      cout << ans << '\n';
    }
  }
  return 0;
}