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 INF = 1e9 + 7;
deque<int> a[318];
int n;
int tot;
int typ, l, r, m;
int g[318][100001];
int ans;
int main() {
  cin >> n;
  int lo = sqrt(n);
  for (int i = 0; i < (n); i++) {
    int t;
    cin >> t;
    a[i / lo].push_back(t);
    g[i / lo][t]++;
  }
  cin >> tot;
  while (tot--) {
    cin >> typ >> l >> r;
    l = ((l + ans - 1) % n) + 1;
    r = ((r + ans - 1) % n) + 1;
    if (l > r) {
      swap(l, r);
    }
    l--;
    r--;
    int ll = l / lo;
    int rr = r / lo;
    if (typ & 1) {
      if (ll == rr) {
        int t = a[ll][r % lo];
        a[ll].erase(a[ll].begin() + r % lo);
        a[ll].insert(a[ll].begin() + l % lo, t);
      } else {
        int p = ll;
        while (p < rr) {
          int t = a[p].back();
          a[p].pop_back();
          g[p][t]--;
          p++;
          a[p].push_front(t);
          g[p][t]++;
        }
        int t = a[rr][r % lo + 1];
        a[rr].erase(a[rr].begin() + r % lo + 1);
        g[rr][t]--;
        a[ll].insert(a[ll].begin() + l % lo, t);
        g[ll][t]++;
      }
    } else {
      cin >> m;
      m = ((m + ans - 1) % n) + 1;
      ans = 0;
      if (ll == rr) {
        for (int i = l % lo; i <= r % lo; i++) {
          if (a[ll][i] == m) {
            ans++;
          }
        }
        cout << ans << endl;
      } else {
        for (int i = ll + 1; i < rr; i++) {
          ans += g[i][m];
        }
        for (int i = l % lo; i < a[ll].size(); i++) {
          if (a[ll][i] == m) {
            ans++;
          }
        }
        for (int i = 0; i <= r % lo; i++) {
          if (a[rr][i] == m) {
            ans++;
          }
        }
        cout << ans << endl;
      }
    }
  }
  return 0;
}