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;
int main() {
  int n, m, s, ca, pre;
  scanf("%d", &n);
  s = sqrt(n);
  int a[n], q, l, r, k, ans = 0, ls, rs;
  m = n / s + 1;
  int b[m][n];
  deque<int> c[n];
  for (int i = 0; i < m; i++)
    for (int j = 0; j < n; j++) b[i][j] = 0;
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  for (int i = 0; i < n; i++) {
    b[i / s][a[i]]++;
    c[i / s].push_back(a[i]);
  }
  scanf("%d", &q);
  while (q--) {
    scanf("%d", &ca);
    if (ca == 1) {
      scanf("%d%d", &l, &r);
      l = 1 + (n + l + ans - 1) % n;
      r = 1 + (n + r + ans - 1) % n;
      if (l > r) swap(l, r);
      l--;
      r--;
      ls = l / s;
      rs = r / s;
      if (ls != rs) {
        pre = c[rs][0];
        for (int i = 0; i <= r % s; i++) swap(pre, c[rs][i]);
        c[rs].pop_front();
        b[ls][pre]++;
        b[rs][pre]--;
        for (int i = l % s; i < c[ls].size(); i++) swap(pre, c[ls][i]);
        b[ls][pre]--;
        for (int i = ls + 1; i < rs; i++) {
          b[i][pre]++;
          c[i].push_front(pre);
          pre = c[i].back();
          c[i].pop_back();
          b[i][pre]--;
        }
        b[rs][pre]++;
        c[rs].push_front(pre);
      } else {
        pre = c[rs][r % s];
        for (int i = l % s; i <= r % s; i++) swap(pre, c[rs][i]);
      }
    } else {
      scanf("%d%d%d", &l, &r, &k);
      l = 1 + (n + l + ans - 1) % n;
      r = 1 + (n + r + ans - 1) % n;
      k = 1 + (n + k + ans - 1) % n;
      if (l > r) swap(l, r);
      ans = 0;
      l--;
      r--;
      ls = l / s;
      rs = r / s;
      if (ls != rs) {
        for (int i = l % s; i < c[ls].size(); i++)
          if (c[ls][i] == k) ans++;
        for (int i = ls + 1; i < rs; i++) ans += b[i][k];
        for (int i = 0; i <= r % s; i++)
          if (c[rs][i] == k) ans++;
      } else {
        for (int i = l % s; i <= r % s; i++)
          if (c[ls][i] == k) ans++;
      }
      printf("%d\n", ans);
    }
  }
}