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 cnt[320][100001];
deque<int> block[320];
int a[100001];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", a + i);
  int lastans = 0, q;
  scanf("%d", &q);
  int sqtn = sqrt(n) + 1;
  for (int i = 0; i < sqtn; i++) {
    for (int j = 0; j < sqtn && i * sqtn + j < n; j++) {
      cnt[i][a[i * sqtn + j]] += 1;
      block[i].push_back(a[i * sqtn + j]);
    }
  }
  while (q--) {
    int type;
    scanf("%d", &type);
    if (type == 1) {
      int _l, _r;
      scanf("%d %d", &_l, &_r);
      int l = (_l + lastans - 1) % n, r = (_r + lastans - 1) % n;
      if (l > r) swap(l, r);
      int lastnum = block[r / sqtn][r % sqtn];
      cnt[l / sqtn][lastnum] += 1;
      cnt[r / sqtn][lastnum] -= 1;
      block[r / sqtn].erase(block[r / sqtn].begin() + r % sqtn);
      block[l / sqtn].insert(block[l / sqtn].begin() + l % sqtn, lastnum);
      if (r < (l / sqtn + 1) * sqtn)
        ;
      else {
        for (int i = 0; i < sqtn; i++) {
          if (l < (i + 1) * sqtn && r >= (i + 1) * sqtn) {
            cnt[i][block[i].back()] -= 1;
            cnt[i + 1][block[i].back()] += 1;
            block[i + 1].push_front(block[i].back());
            block[i].pop_back();
          }
        }
      }
    } else if (type == 2) {
      int _l, _r, _k;
      scanf("%d %d %d", &_l, &_r, &_k);
      int l, r, k;
      l = (_l + lastans - 1) % n;
      r = (_r + lastans - 1) % n;
      k = (_k + lastans - 1) % n + 1;
      if (l > r) swap(l, r);
      lastans = 0;
      for (int base = 0, i = 0; base < n; ++i, base += sqtn) {
        if (base + sqtn <= l || r + 1 <= base) continue;
        if (l <= base && base + sqtn <= r + 1)
          lastans += cnt[i][k];
        else
          for (int j = 0; j < sqtn && i * sqtn + j < n; ++j) {
            int idx = base + j;
            if (l <= idx && idx < r + 1) {
              if (block[i][idx % sqtn] == k) lastans += 1;
            }
          }
      }
      printf("%d\n", lastans);
    }
  }
}