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
#include <bits/stdc++.h>
using namespace std;
const int block = 320;
int cnt[block][100005];
deque<int> baris[block];
int main() {
  int n, q, a, l, r, block1, block2, urutan1, urutan2, ans, stat, tampung, k;
  ans = 0;
  scanf("%d", &n);
  baris[0].push_back(0);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a);
    cnt[i / block][a]++;
    baris[i / block].push_back(a);
  }
  scanf("%d", &q);
  for (int i = 0; i < q; i++) {
    scanf("%d %d %d", &stat, &l, &r);
    l = (l + ans - 1) % n + 1;
    r = (r + ans - 1) % n + 1;
    if (l > r) {
      int tmp = l;
      l = r;
      r = tmp;
    }
    block1 = l / block;
    urutan1 = l % block;
    block2 = r / block;
    urutan2 = r % block;
    if (stat == 1) {
      tampung = baris[block2][urutan2];
      baris[block2].erase(baris[block2].begin() + urutan2);
      baris[block1].insert(baris[block1].begin() + urutan1, tampung);
      if (block1 == block2) {
      } else {
        cnt[block2][tampung]--;
        cnt[block1][tampung]++;
        for (int j = block1; j < block2; j++) {
          tampung = baris[j].back();
          cnt[j][tampung]--;
          cnt[j + 1][tampung]++;
          baris[j].pop_back();
          baris[j + 1].push_front(tampung);
        }
      }
    } else {
      scanf("%d", &k);
      k = (k + ans - 1) % n + 1;
      ans = 0;
      if (block1 == block2) {
        for (int j = urutan1; j <= urutan2; j++) {
          if (baris[block1][j] == k) {
            ans++;
          }
        }
      } else {
        for (int j = urutan1; j < block; j++) {
          if (baris[block1][j] == k) ans++;
        }
        for (int j = 0; j <= urutan2; j++) {
          if (baris[block2][j] == k) ans++;
        }
        for (int j = block1 + 1; j < block2; j++) {
          ans += cnt[j][k];
        }
      }
      printf("%d\n", ans);
    }
  }
  return 0;
}
