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 MAXN = 1e5 + 10000;
const int SQ = 450;
deque<int> deq[MAXN / SQ];
int mp[MAXN / SQ][MAXN];
int a[MAXN];
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++)
    cin >> a[i], deq[i / SQ].push_back(a[i]), mp[i / SQ][a[i]]++;
  int q;
  cin >> q;
  int lastans = 0;
  while (q--) {
    int qt;
    cin >> qt;
    if (qt == 1) {
      int l, r;
      cin >> l >> r;
      r = (r + lastans - 1) % n;
      l = (l + lastans - 1) % n;
      if (r < l) swap(r, l);
      int temp = deq[r / SQ][r % SQ];
      mp[l / SQ][temp]++;
      mp[r / SQ][temp]--;
      deq[r / SQ].erase(deq[r / SQ].begin() + r % SQ);
      deq[l / SQ].insert(deq[l / SQ].begin() + l % SQ, temp);
      l /= SQ;
      r /= SQ;
      for (int i = l; i < r; i++) {
        int x = deq[i].back();
        mp[i + 1][x]++;
        deq[i + 1].push_front(x);
        mp[i][x]--;
        deq[i].pop_back();
      }
    } else {
      int l, r, k;
      cin >> l >> r >> k;
      r = (r + lastans - 1) % n;
      l = (l + lastans - 1) % n;
      k = (k + lastans - 1) % n + 1;
      if (r < l) swap(r, l);
      r++;
      lastans = 0;
      while (r % SQ) {
        r--;
        lastans += (deq[r / SQ][r % SQ] == k);
      }
      while (l % SQ) {
        l--;
        lastans -= (deq[l / SQ][l % SQ] == k);
      }
      for (int i = l; i < r; i += SQ) lastans += mp[i / SQ][k];
      cout << lastans << endl;
    }
  }
}