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 C = 350, MAXN = 1e5 + C - 1, MAXC = MAXN / C;
int cnt[MAXC][MAXN];
deque<int> deq[MAXC];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    int c = i / C, a;
    scanf("%d", &a);
    deq[c].push_back(a);
    cnt[c][a]++;
  }
  int q;
  cin >> q;
  int lastans = 0;
  while (q--) {
    int t, l, r;
    scanf("%d%d%d", &t, &l, &r);
    l = ((l + lastans - 1) % n) + 1, r = ((r + lastans - 1) % n) + 1;
    if (l > r) swap(l, r);
    l--;
    if (t == 1) {
      int add = deq[(r - 1) / C][r - (r - 1) / C * C - 1];
      for (int s = l / C; s * C < r; s++) {
        int sl = s * C, sr = sl + C;
        if (l <= sl && r >= sr) {
          cnt[s][add]++;
          deq[s].push_front(add);
          add = deq[s].back();
          cnt[s][add]--;
          deq[s].pop_back();
        } else {
          int L = max(sl, l), R = min(sr, r), tmp = deq[s][R - 1 - sl];
          cnt[s][add]++;
          cnt[s][tmp]--;
          for (int i = R - 1 - sl; i > L - sl; i--) deq[s][i] = deq[s][i - 1];
          deq[s][L - sl] = add;
          add = tmp;
        }
      }
    } else {
      int k;
      scanf("%d", &k);
      k = ((k + lastans - 1) % n) + 1;
      int ans = 0;
      for (int s = l / C; s * C < r; s++) {
        int sl = s * C, sr = sl + C;
        if (l <= sl && r >= sr)
          ans += cnt[s][k];
        else {
          int L = max(l, sl), R = min(r, sr);
          for (int i = L; i < R; i++)
            if (deq[s][i - sl] == k) ans++;
        }
      }
      printf("%d\n", ans);
      lastans = ans;
    }
  }
}