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 N = 100010;
const int M = 400;
deque<int> d[M];
int cnt[M][N], a[N];
int main(void) {
  int n;
  scanf("%d", &n);
  int size_block = n / sqrt(n);
  for (int i = 0; i < n; i++) {
    scanf("%d", a + i);
    d[i / size_block].push_back(a[i]);
    cnt[i / size_block][a[i]]++;
  }
  int q;
  scanf("%d", &q);
  int last_ans = 0;
  while (q--) {
    int t, l, r;
    scanf("%d %d %d", &t, &l, &r);
    l = (l + last_ans - 1) % n + 1;
    r = (r + last_ans - 1) % n + 1;
    l--;
    r--;
    if (l > r) swap(l, r);
    int a = l / size_block, b = r / size_block;
    if (t == 1) {
      if (a == b) {
        int x = a * size_block, y = min(n, (a + 1) * size_block);
        deque<int> aux;
        int guarda;
        for (int j = y - 1; j >= l; j--) {
          if (j != r)
            aux.push_front(d[a].back());
          else
            guarda = d[a].back();
          d[a].pop_back();
        }
        d[a].push_back(guarda);
        while (!aux.empty()) {
          d[a].push_back(aux.front());
          aux.pop_front();
        }
      } else {
        int x = b * size_block, y = (b + 1) * size_block;
        deque<int> aux;
        for (int i = x; i <= r; i++) {
          aux.push_back(d[b].front());
          d[b].pop_front();
        }
        int guarda = aux.back();
        aux.pop_back();
        while (!aux.empty()) {
          d[b].push_front(aux.back());
          aux.pop_back();
        }
        cnt[b][guarda]--;
        x = a * size_block, y = (a + 1) * size_block;
        for (int i = y - 1; i >= l; i--) {
          aux.push_front(d[a].back());
          d[a].pop_back();
        }
        d[a].push_back(guarda);
        cnt[a][guarda]++;
        while (!aux.empty()) {
          d[a].push_back(aux.front());
          aux.pop_front();
        }
        guarda = d[a].back();
        d[a].pop_back();
        cnt[a][guarda]--;
        for (int i = a + 1; i < b; i++) {
          d[i].push_front(guarda);
          cnt[i][guarda]++;
          guarda = d[i].back();
          d[i].pop_back();
          cnt[i][guarda]--;
        }
        d[b].push_front(guarda);
        cnt[b][guarda]++;
      }
    } else {
      int k;
      scanf("%d", &k);
      k = (k + last_ans - 1) % n + 1;
      int ans = 0;
      int x = a * size_block, y = (a + 1) * size_block;
      for (int i = l; i < min(y, r + 1); i++) ans += d[a][i - x] == k ? 1 : 0;
      for (int i = a + 1; i < b; i++) ans += cnt[i][k];
      if (a != b) {
        x = b * size_block, y = (b + 1) * size_block;
        for (int i = x; i <= r; i++) ans += d[b][i - x] == k ? 1 : 0;
      }
      cout << ans << endl;
      last_ans = ans;
    }
  }
  return 0;
}
