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 = 100005;
deque<int> a[350];
int n, q, s, ans, c[350][N];
inline void deal(int &x) { x = (x + ans - 1) % n + 1; }
int main() {
  scanf("%d", &n);
  s = sqrt(n);
  for (int i = 1, x; i <= n; ++i) {
    scanf("%d", &x);
    a[(i - 1) / s + 1].push_back(x);
    c[(i - 1) / s + 1][x]++;
  }
  scanf("%d", &q);
  int op, l, r, x;
  while (q--) {
    scanf("%d%d%d", &op, &l, &r);
    deal(l), deal(r);
    if (l > r) swap(l, r);
    int fl = (l - 1) / s + 1, fr = (r - 1) / s + 1;
    if (op == 1) {
      if (fl == fr) {
        int p = (r - 1) % s;
        x = a[fl][p];
        a[fl].erase(a[fl].begin() + p);
        a[fl].insert(a[fl].begin() + (l - 1) % s, x);
      } else {
        for (int i = fl; i < fr; ++i) {
          x = a[i].back(), a[i].pop_back(), --c[i][x];
          a[i + 1].push_front(x), c[i + 1][x]++;
        }
        int p = (r - 1) % s + 1;
        x = a[fr][p];
        a[fr].erase(a[fr].begin() + p);
        --c[fr][x];
        a[fl].insert(a[fl].begin() + (l - 1) % s, x);
        ++c[fl][x];
      }
    } else {
      scanf("%d", &x);
      deal(x);
      ans = 0;
      if (fl == fr)
        for (int i = (l - 1) % s; i <= (r - 1) % s; ++i) ans += (a[fl][i] == x);
      else {
        for (int i = (l - 1) % s; i < s; ++i) ans += (a[fl][i] == x);
        for (int i = fl + 1; i < fr; ++i) ans += c[i][x];
        for (int i = 0; i <= (r - 1) % s; ++i) ans += (a[fr][i] == x);
      }
      printf("%d\n", ans);
    }
  }
  return 0;
}