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 long long int mx = 1e5 + 1, Mod = 1e9 + 7, mst = 320;
int n, q, ls, st = 316;
deque<int> seg[mst];
int cnt[mst][mx];
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n;
  int qt, l, r, k, nod, tmp;
  for (int i = 0; i < n; i++)
    cin >> k, seg[i / st].push_back(k), cnt[i / st][k]++;
  cin >> q;
  while (q--) {
    cin >> qt >> l >> r;
    l = (l + ls - 1 + n) % n;
    r = (r + ls - 1 + n) % n;
    if (r < l) swap(l, r);
    if (qt == 1) {
      if (l / st == r / st) {
        nod = r / st;
        for (int i = r % st - 1; i >= l % st; i--)
          swap(seg[nod][i], seg[nod][i + 1]);
      } else {
        for (nod = l / st; nod < r / st; nod++) {
          tmp = seg[nod][seg[nod].size() - 1];
          seg[nod].pop_back();
          cnt[nod][tmp]--;
          seg[nod + 1].push_front(tmp);
          cnt[nod + 1][tmp]++;
        }
        nod = r / st;
        for (int i = r % st; i >= 0; i--) swap(seg[nod][i], seg[nod][i + 1]);
        tmp = seg[nod][0];
        seg[nod].pop_front();
        cnt[nod][tmp]--;
        nod = l / st;
        seg[nod].push_back(tmp);
        cnt[nod][tmp]++;
        for (int i = seg[nod].size() - 2; i >= l % st; i--)
          swap(seg[nod][i], seg[nod][i + 1]);
      }
    } else {
      cin >> k;
      k = (k + ls - 1 + n) % n + 1;
      ls = 0;
      if (l / st == r / st) {
        for (int i = l % st; i <= r % st; i++) ls += (seg[l / st][i] == k);
      } else {
        for (int i = l / st + 1; i < r / st; i++) ls += cnt[i][k];
        for (int i = l % st; i < seg[l / st].size(); i++)
          ls += (seg[l / st][i] == k);
        for (int i = 0; i <= r % st; i++) ls += (seg[r / st][i] == k);
      }
      cout << ls << '\n';
    }
  }
  return 0;
}