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 ans = 0;
deque<int> dq[400];
int buc[350][100005];
int main() {
  int n;
  scanf("%d", &n);
  int div = sqrt(n);
  memset(buc, 0, sizeof(buc));
  for (int i = 0; i < n; i++) {
    int x;
    scanf("%d", &x);
    dq[i / div].push_back(x);
    buc[i / div][x]++;
  }
  int q;
  scanf("%d", &q);
  while (q--) {
    int typ, l, r;
    scanf("%d%d%d", &typ, &l, &r);
    l = (l + ans - 1) % n + 1;
    r = (r + ans - 1) % n + 1;
    if (l > r) swap(l, r);
    l--;
    int l1 = l / div, r1 = r / div, l2 = l % div, r2 = r % div;
    if (typ == 1) {
      if (l1 == r1) {
        r2--;
        int x = dq[r1][r2];
        dq[r1].erase(dq[r1].begin() + r2);
        dq[l1].insert(dq[l1].begin() + l2, x);
      } else {
        for (int i = l1; i < r1; i++) {
          int x = dq[i].back();
          dq[i].pop_back();
          buc[i][x]--;
          dq[i + 1].push_front(x);
          buc[i + 1][x]++;
        }
        int x = dq[r1][r2];
        dq[r1].erase(dq[r1].begin() + r2);
        buc[r1][x]--;
        dq[l1].insert(dq[l1].begin() + l2, x);
        buc[l1][x]++;
      }
    } else if (typ == 2) {
      int k;
      scanf("%d", &k);
      k = (k + ans - 1) % n + 1;
      ans = 0;
      if (l1 == r1)
        for (int i = l2; i < r2; i++) ans += (dq[l1][i] == k);
      else {
        for (int i = l1 + 1; i < r1; i++) ans += buc[i][k];
        for (int i = l2; i < dq[l1].size(); i++) ans += (dq[l1][i] == k);
        for (int i = 0; i < r2; i++) ans += (dq[r1][i] == k);
      }
      printf("%d\n", ans);
    }
  }
  return 0;
}