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
#include <bits/stdc++.h>
using namespace std;
int a[100005];
deque<int> v[320];
int n, q, S, res;
int cnt[330][100005];
void update(int l, int r) {
  int cl = l / S, cr = r / S;
  int b = v[cl][l % S];
  int c = v[cr][r % S];
  if (cl == cr) {
    v[cl].erase(v[cl].begin() + r % S);
    v[cl].insert(v[cl].begin() + l % S, c);
  } else {
    v[cr].erase(v[cr].begin() + r % S);
    cnt[cr][c]--;
    v[cl].insert(v[cl].begin() + l % S, c);
    cnt[cl][c]++;
    for (int i = cl + 1; i <= cr; i++) {
      int d = v[i - 1][v[i - 1].size() - 1];
      v[i].push_front(d);
      cnt[i][d]++;
      v[i - 1].pop_back();
      cnt[i - 1][d]--;
    }
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(NULL);
  cin >> n;
  S = sqrt(n * 1.0) + 1;
  v[0].push_back(0);
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    v[i / S].push_back(a[i]);
    cnt[i / S][a[i]]++;
  }
  int q;
  cin >> q;
  for (int i = 1; i <= q; i++) {
    int type;
    cin >> type;
    if (type == 1) {
      int l, r;
      cin >> l >> r;
      l = (l + res - 1) % n + 1;
      r = (r + res - 1) % n + 1;
      if (l > r) swap(l, r);
      update(l, r);
    } else {
      int l, r, k;
      cin >> l >> r >> k;
      l = (l + res - 1) % n + 1;
      r = (r + res - 1) % n + 1;
      k = (k + res - 1) % n + 1;
      if (l > r) swap(l, r);
      int cl = l / S;
      int cr = r / S;
      int ans = 0;
      if (cl == cr) {
        for (int i = l; i <= r; i++) {
          if (v[i / S][i % S] == k) ans++;
        }
      } else {
        for (int i = l; i <= (cl + 1) * S - 1; i++) {
          if (v[i / S][i % S] == k) ans++;
        }
        for (int i = cr * S; i <= r; i++) {
          if (v[i / S][i % S] == k) ans++;
        }
        for (int i = cl + 1; i <= cr - 1; i++) {
          ans += cnt[i][k];
        }
      }
      res = ans;
      cout << res << endl;
    }
  }
  return 0;
}
