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
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e5 + 10, SQ = 400;
deque<int> v[SQ];
int a[MAXN], b[SQ][MAXN];
int main() {
  int n, cnt = -1;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", a + i);
    if (i % SQ == 0) cnt++;
    v[cnt].push_back(a[i]);
    b[cnt][a[i]]++;
  }
  int q, lastans = 0;
  scanf("%d", &q);
  for (int i = 0, qt; i < q; i++) {
    scanf("%d", &qt);
    if (qt == 1) {
      int l, r;
      scanf("%d%d", &l, &r);
      l = ((l + lastans - 1) % n);
      r = ((r + lastans - 1) % n);
      if (l > r) swap(l, r);
      int tmp = v[r / SQ][r % SQ];
      b[r / SQ][tmp]--;
      b[l / SQ][tmp]++;
      v[r / SQ].erase(v[r / SQ].begin() + r % SQ);
      v[l / SQ].insert(v[l / SQ].begin() + l % SQ, tmp);
      for (int j = l / SQ + 1; j <= r / SQ; j++) {
        b[j][v[j - 1].back()]++;
        b[j - 1][v[j - 1].back()]--;
        v[j].push_front(v[j - 1].back());
        v[j - 1].pop_back();
      }
    }
    if (qt == 2) {
      int l, r, val, ans = 0;
      scanf("%d%d%d", &l, &r, &val);
      l = ((l + lastans - 1) % n) + 1;
      r = ((r + lastans - 1) % n) + 1;
      if (l > r) swap(l, r);
      l--;
      val = ((val + lastans - 1) % n) + 1;
      while (l < r && l % SQ != 0) ans += (v[l / SQ][l % SQ] == val), l++;
      while (r > l && r % SQ != 0) r--, ans += (v[r / SQ][r % SQ] == val);
      for (int j = l / SQ; j < r / SQ; j++) ans += b[j][val];
      printf("%d\n", ans);
      lastans = ans;
    }
  }
  return 0;
}
