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 f[320][100086];
int n, sn;
int opt, l, r, k;
int m, x;
deque<int> q[320];
int lastans;
int main() {
  scanf("%d", &n), sn = (int)sqrt(n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &x);
    int ii = (i - 1) / sn;
    q[ii].push_back(x), ++f[ii][x];
  }
  scanf("%d", &m);
  while (m--) {
    scanf("%d", &opt);
    if (opt == 1) {
      scanf("%d%d", &l, &r);
      l = (l + lastans - 1) % n + 1, r = (r + lastans - 1) % n + 1;
      if (l > r) swap(l, r);
      int ll = (l - 1) / sn, rr = (r - 1) / sn, li = (l - 1) % sn,
          ri = (r - 1) % sn;
      if (ll == rr) {
        int x = q[ll][ri];
        q[ll].erase(q[ll].begin() + ri);
        q[ll].insert(q[ll].begin() + li, x);
      } else {
        int x = q[ll].back();
        q[ll].pop_back(), --f[ll][x];
        for (int i = ll + 1; i < rr; i++) {
          q[i].push_front(x), ++f[i][x];
          x = q[i].back(), q[i].pop_back(), --f[i][x];
        }
        q[rr].push_front(x), ++f[rr][x];
        x = q[rr][ri + 1], --f[rr][x], q[rr].erase(q[rr].begin() + ri + 1);
        ++f[ll][x], q[ll].insert(q[ll].begin() + li, x);
      }
    } else {
      scanf("%d%d%d", &l, &r, &k);
      l = (l + lastans - 1) % n + 1, r = (r + lastans - 1) % n + 1,
      k = (k + lastans - 1) % n + 1;
      if (l > r) swap(l, r);
      int ll = (l - 1) / sn, rr = (r - 1) / sn, li = (l - 1) % sn,
          ri = (r - 1) % sn;
      int ans = 0;
      if (ll == rr) {
        for (int i = li; i <= ri; i++) ans += q[ll][i] == k;
      } else {
        for (int i = ll + 1; i < rr; i++) ans += f[i][k];
        for (int i = li; i < sn; i++) ans += q[ll][i] == k;
        for (int i = 0; i <= ri; i++) ans += q[rr][i] == k;
      }
      printf("%d\n", lastans = ans);
    }
  }
}