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 a[100000];
int n;
int tot = 0;
struct block {
  int cnt[100001], sz, p[700];
  void init(int _sz, int *_p) {
    sz = _sz;
    memset(cnt, 0, sizeof(cnt));
    for (int i = 0; i < sz; i++) cnt[p[i] = _p[i]]++;
  }
  int del(int x) {
    int tmp = p[x];
    cnt[tmp]--;
    sz--;
    for (int i = x; i < sz; i++) p[i] = p[i + 1];
    return tmp;
  }
  void insert(int x, int val) {
    sz++;
    for (int i = sz - 1; i > x; i--) p[i] = p[i - 1];
    p[x] = val;
    cnt[val]++;
  }
  int sum(int l, int r, int x) {
    if (r <= 0 || l >= sz) return 0;
    if (l <= 0 && r >= sz)
      return cnt[x];
    else {
      int res = 0;
      for (int i = max(0, l); i < r && i < sz; i++) res += (p[i] == x);
      return res;
    }
  }
} b[500];
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n; i += 333) b[tot++].init(min(n, i + 333) - i, a + i);
  int res = 0, op, l, r, t, q = 0;
  cin >> t;
  for (int k = 0; k < t; k++) {
    cin >> op;
    cin >> l >> r;
    l = ((l + res - 1)) % n;
    r = ((r + res - 1)) % n;
    if (l > r) swap(l, r);
    if (op == 1) {
      int tmp;
      for (int x = 0, i = 0; i < tot; x += b[i++].sz)
        if (r - x < b[i].sz && r - x >= 0) {
          tmp = b[i].del(r - x);
          break;
        }
      for (int x = 0, i = 0; i < tot; x += b[i++].sz)
        if (l - x >= 0 && l - x <= b[i].sz) {
          b[i].insert(l - x, tmp);
          break;
        }
      if ((++q) == 333) {
        for (int x = 0, i = 0; i < tot; x += b[i++].sz)
          for (int j = 0; j < b[i].sz; j++) a[x + j] = b[i].p[j];
        q = tot = 0;
        for (int i = 0; i < n; i += 333)
          b[tot++].init(min(n, i + 333) - i, a + i);
      }
    } else {
      int tmp;
      cin >> tmp;
      tmp = ((tmp + res - 1) % n) + 1;
      res = 0;
      r++;
      for (int i = 0, x = 0; i < tot; x += b[i++].sz)
        res += b[i].sum(l - x, r - x, tmp);
      cout << res << endl;
    }
  }
  return 0;
}