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 + 21;
const int zn = 351;
int n, q, ans, dp[zn][maxn];
deque<int> Q[zn];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) {
    int a;
    cin >> a;
    a--;
    dp[i / zn][a]++;
    Q[i / zn].push_back(a);
  }
  cin >> q;
  for (int i = 0; i < q; i++) {
    int t, l, r;
    cin >> t >> l >> r;
    l = (l + ans - 1) % n;
    r = (r + ans - 1) % n;
    if (l > r) swap(l, r);
    int Lzone = l / zn, Rzone = r / zn;
    int Lmod = l % zn, Rmod = r % zn;
    if (t == 2) {
      int k;
      cin >> k;
      k = (k + ans - 1) % n;
      ans = 0;
      if (Lzone == Rzone) {
        for (int j = Lmod; j <= Rmod; j++)
          if (Q[Lzone][j] == k) ans++;
        cout << ans << "\n";
        continue;
      }
      for (int j = Lmod; j < zn; j++)
        if (Q[Lzone][j] == k) ans++;
      Lzone++;
      while (Lzone != Rzone) {
        ans += dp[Lzone][k];
        Lzone++;
      }
      for (int j = 0; j <= Rmod; j++)
        if (Q[Rzone][j] == k) ans++;
      cout << ans << "\n";
    } else {
      int x = Q[Rzone][Rmod];
      Q[Rzone].erase(Q[Rzone].begin() + (Rmod));
      dp[Rzone][x]--;
      Q[Lzone].insert(Q[Lzone].begin() + (Lmod), x);
      dp[Lzone][x]++;
      while (Lzone != Rzone) {
        x = Q[Lzone].back();
        Q[Lzone].pop_back();
        dp[Lzone][x]--;
        Q[Lzone + 1].push_front(x);
        dp[Lzone + 1][x]++;
        Lzone++;
      }
    }
  }
}