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;
const long long S = 320;
deque<long long> v[S];
long long b[S][100100], ls;
int main() {
  long long n;
  cin >> n;
  for (long long i = 0; i < n; i++) {
    long long x;
    cin >> x;
    v[i / S].push_back(x);
    b[i / S][x]++;
  }
  long long q;
  cin >> q;
  while (q--) {
    long long tp, l, r;
    cin >> tp >> l >> r;
    l = (l + ls - 1) % n;
    r = (r + ls - 1) % n;
    if (l > r) swap(l, r);
    if (tp - 1) {
      long long k, ans = 0;
      cin >> k;
      k = ((k + ls - 1) % n) + 1;
      if (l / S == r / S) {
        for (long long i = l % S; i <= r % S; i++)
          if (v[l / S][i] == k) ans++;
      } else {
        for (long long i = l % S; i < v[l / S].size(); i++)
          if (v[l / S][i] == k) ans++;
        for (long long i = l / S + 1; i < r / S; i++) ans += b[i][k];
        for (long long i = 0; i <= r % S; i++)
          if (v[r / S][i] == k) ans++;
      }
      cout << (ls = ans) << endl;
    } else {
      b[l / S][v[r / S][r % S]]++;
      v[l / S].push_front(v[r / S][r % S]);
      for (long long i = 0; i < l % S; i++) swap(v[l / S][i], v[l / S][i + 1]);
      for (long long i = l / S; i < r / S; i++)
        b[i + 1][v[i].back()]++, v[i + 1].push_front(v[i].back()),
            b[i][v[i].back()]--, v[i].pop_back();
      for (long long i = r % S + 1; i < v[r / S].size() - 1; i++)
        swap(v[r / S][i], v[r / S][i + 1]);
      b[r / S][v[r / S].back()]--;
      v[r / S].pop_back();
    }
  }
  return 0;
}