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>
#pragma GCC optimize("O2,unroll-loops")
#pragma GCC target("avx,avx2")
inline int read() {
  int x = 0;
  char c = getchar();
  while (!isdigit(c)) c = getchar();
  while (isdigit(c)) x = (x << 3) + (x << 1) + c - 48, c = getchar();
  return x;
}
const int N = 1e5 + 7;
int n, _, sz, a[N];
int main() {
  n = read();
  for (int i = 1; i <= n; ++i) a[i] = read();
  _ = read();
  int ans = 0;
  while (_-- > 0) {
    int op = read(), l = read(), r = read();
    l = ((l + ans - 1) % n) + 1;
    r = ((r + ans - 1) % n) + 1;
    if (l > r) std::swap(l, r);
    if (op == 1) {
      int t = a[r];
      while (r > l) a[r] = a[r - 1], --r;
      a[l] = t;
    } else {
      int k = read();
      k = ((k + ans - 1) % n) + 1;
      int ret = 0;
      for (int i = l; i <= r; ++i) ret += a[i] == k;
      printf("%d\n", ans = ret);
    }
  }
  return 0;
}