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 int N = 1e5 + 5, K = 333;
deque<int> d[K];
int buc[K][N], ans, n, q;
int cal(int x) { return (x - 1) / K; }
int pos(int x) { return (x - 1) % K; }
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x;
    int id = cal(i);
    d[id].push_back(x);
    buc[id][x]++;
  }
  cin >> q;
  for (int i = 1; i <= q; i++) {
    int op, l, r;
    cin >> op >> l >> r;
    l = (l + ans - 1) % n + 1, r = (r + ans - 1) % n + 1;
    if (l > r) swap(l, r);
    if (op == 1) {
      int idl = cal(l), idr = cal(r), pl = pos(l), pr = pos(r);
      int val = d[idr][pr];
      d[idr].erase(d[idr].begin() + pr);
      buc[idr][val]--;
      d[idl].insert(d[idl].begin() + pl, val);
      buc[idl][val]++;
      for (int j = idl; j < idr; j++) {
        val = d[j].back();
        d[j].pop_back();
        buc[j][val]--;
        d[j + 1].push_front(val);
        buc[j + 1][val]++;
      }
    } else {
      int k, idl = cal(l), idr = cal(r), pl = pos(l), pr = pos(r);
      cin >> k;
      k = (k + ans - 1) % n + 1;
      ans = 0;
      if (idl == idr)
        for (int j = pl; j <= pr; j++) ans += d[idl][j] == k;
      else {
        for (int j = pl; j < K; j++) ans += d[idl][j] == k;
        for (int j = 0; j <= pr; j++) ans += d[idr][j] == k;
        for (int j = idl + 1; j < idr; j++) ans += buc[j][k];
      }
      printf("%d\n", ans);
    }
  }
  return 0;
}