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;
signed main() {
  std::ios_base::sync_with_stdio(0), cin.tie(0);
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < n; ++i) cin >> a[i];
  vector<vector<int>> bl((n + 511) / 512);
  vector<unordered_map<int, int>> kol((n + 511) / 512);
  int j = -1;
  for (int i = 0; i < n; ++i) {
    if (i % 512 == 0) j++;
    bl[j].push_back(a[i]);
    kol[j][a[i]]++;
  }
  int q;
  cin >> q;
  int ty1 = 0;
  int last = 0;
  while (q--) {
    if (ty1 == 512) {
      int cnt = 0;
      for (int i = 0; i < bl.size(); ++i) {
        for (int h = 0; h < bl[i].size(); ++h) a[cnt] = bl[i][h], cnt++;
        bl[i].clear();
        kol[i].clear();
      }
      j = -1;
      for (int i = 0; i < n; ++i) {
        if (i % 512 == 0) j++;
        bl[j].push_back(a[i]);
        kol[j][a[i]]++;
      }
      ty1 = 0;
    }
    int ans = 0;
    int tp;
    cin >> tp;
    if (tp == 1) {
      int l, r;
      cin >> l >> r;
      l = (l + last - 1) % n + 1, r = (r + last - 1) % n + 1;
      if (l > r) swap(l, r);
      l--, r--;
      int cnt = 0;
      int ins = 0;
      for (int i = 0; i < bl.size(); ++i) {
        if (r >= cnt && r < cnt + bl[i].size()) {
          r -= cnt;
          ins = bl[i][r];
          for (int h = r; h < (int)bl[i].size() - 1; ++h)
            bl[i][h] = bl[i][h + 1];
          bl[i].pop_back();
          kol[i][ins]--;
          break;
        }
        cnt += bl[i].size();
      }
      cnt = 0;
      for (int i = 0; i < bl.size(); ++i) {
        if (l >= cnt && l < cnt + bl[i].size()) {
          bl[i].push_back(0);
          l -= cnt;
          for (int h = (int)bl[i].size() - 1; h > l; --h)
            bl[i][h] = bl[i][h - 1];
          bl[i][l] = ins;
          kol[i][ins]++;
          break;
        }
        cnt += bl[i].size();
      }
      ty1++;
    } else {
      int l, r, k;
      cin >> l >> r >> k;
      l = (l + last - 1) % n + 1, r = (r + last - 1) % n + 1,
      k = (k + last - 1) % n + 1;
      l--, r--;
      if (l > r) swap(l, r);
      int cnt = 0;
      for (int i = 0; i < bl.size(); ++i) {
        if (l <= cnt && cnt + (int)bl[i].size() - 1 <= r) {
          ans += kol[i][k];
        } else if (l >= cnt && l < cnt + bl[i].size()) {
          for (int h = l - cnt; h <= min(r - cnt, (int)bl[i].size() - 1); ++h)
            if (bl[i][h] == k) ans++;
        } else if (r >= cnt && r < cnt + bl[i].size()) {
          for (int h = 0; h <= r - cnt; ++h)
            if (bl[i][h] == k) ans++;
        }
        cnt += bl[i].size();
      }
      cout << ans << '\n';
      last = ans;
    }
  }
  return 0;
}