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;
const int N = 1e5 + 9, B = 320;
int a[N];
int num[B][N];
deque<int> part[B];
void EF(int ind) { num[ind][part[ind].front()]--, part[ind].pop_front(); }
void EB(int ind) { num[ind][part[ind].back()]--, part[ind].pop_back(); }
void AF(int ind, int val) { num[ind][val]++, part[ind].push_front(val); }
void AB(int ind, int val) { num[ind][val]++, part[ind].push_back(val); }
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], AB(i / B, a[i]);
  while (part[(n - 1) / B].size() < B) AB((n - 1) / B, 0);
  vector<int> vec;
  int q;
  cin >> q;
  int prv = 0;
  while (q--) {
    int f;
    cin >> f;
    if (f == 1) {
      int l, r;
      cin >> l >> r;
      l = ((l + prv - 1) % n) + 1;
      r = ((r + prv - 1) % n) + 1;
      if (l >= r) swap(l, r);
      l--;
      r--;
      int rr = r / B, ll = l / B;
      int val;
      for (int i = 0; i < r % B; i++)
        vec.push_back(part[rr].front()), part[rr].pop_front();
      val = part[rr].front();
      part[rr].pop_front();
      num[rr][val]--;
      for (int i = vec.size() - 1; i >= 0; i--) part[rr].push_front(vec[i]);
      vec.clear();
      for (int i = rr; i > ll; i--) {
        int x = part[i - 1].back();
        EB(i - 1);
        AF(i, x);
      }
      for (int i = 0; i < B - (l % B) - 1; i++)
        vec.push_back(part[ll].back()), part[ll].pop_back();
      AB(ll, val);
      for (int i = vec.size() - 1; i >= 0; i--) part[ll].push_back(vec[i]);
      vec.clear();
    }
    if (f == 2) {
      int l, r, k;
      cin >> l >> r >> k;
      l = ((l + prv - 1) % n) + 1;
      r = ((r + prv - 1) % n) + 1;
      k = ((k + prv - 1) % n) + 1;
      if (l >= r) swap(l, r);
      l--;
      r--;
      int rr = r / B, ll = l / B;
      int ans = 0;
      if (ll == rr) {
        while (!part[ll].empty())
          vec.push_back(part[ll].front()), part[ll].pop_front();
        for (int i = l % B; i <= r % B; i++) ans += (vec[i] == k);
        for (int i = 0; i < B; i++) part[ll].push_back(vec[i]);
        vec.clear();
      } else {
        for (int i = ll + 1; i < rr; i++) ans += num[i][k];
        for (int i = 0; i <= r % B; i++)
          vec.push_back(part[rr].front()), part[rr].pop_front();
        for (int i = vec.size() - 1; i >= 0; i--)
          ans += (vec[i] == k), part[rr].push_front(vec[i]);
        vec.clear();
        for (int i = 0; i < B - (l % B); i++)
          vec.push_back(part[ll].back()), part[ll].pop_back();
        for (int i = vec.size() - 1; i >= 0; i--)
          ans += (vec[i] == k), part[ll].push_back(vec[i]);
        vec.clear();
      }
      cout << ans << '\n';
      prv = ans;
    }
  }
  return 0;
}
