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 N = 1e5 + 11;
const long long mod = 1e9 + 7;
const int inf = 1e9;
const int bl = 320;
deque<int> arr[bl];
int cnt[bl][N];
void shift(int l, int r) {
  vector<int> a;
  int b = l / bl;
  while (!arr[b].empty()) {
    a.push_back(arr[b].front());
    arr[b].pop_front();
  }
  int last = a[r - b * bl];
  for (int i = l - b * bl; i <= r - b * bl; i++) {
    swap(a[i], last);
  }
  while (!a.empty()) {
    arr[b].push_front(a.back());
    a.pop_back();
  }
}
int del(int first) {
  int b = first / bl;
  int second = first - b * bl;
  vector<int> a;
  for (int i = 0; i < second; i++) {
    a.push_back(arr[b].front());
    arr[b].pop_front();
  }
  int res = arr[b].front();
  arr[b].pop_front();
  while (!a.empty()) {
    arr[b].push_front(a.back());
    a.pop_back();
  }
  return res;
}
void add(int first, int val) {
  int b = first / bl;
  int second = first - b * bl;
  vector<int> a;
  for (int i = arr[b].size() - 1; i >= second; i--) {
    a.push_back(arr[b].back());
    arr[b].pop_back();
  }
  arr[b].push_back(val);
  while (!a.empty()) {
    arr[b].push_back(a.back());
    a.pop_back();
  }
}
void solve() {
  int n;
  cin >> n;
  arr[0].push_back(0);
  for (int i = 1; i <= n; i++) {
    int first;
    cin >> first;
    arr[i / bl].push_back(first);
    cnt[i / bl][first]++;
  }
  int q;
  cin >> q;
  int lastans = 0;
  for (int i = 1; i <= q; i++) {
    int type, l, r, k;
    cin >> type >> l >> r;
    if (type == 1) {
      l = (l + lastans - 1) % n + 1;
      r = (r + lastans - 1) % n + 1;
      if (l > r) swap(l, r);
      if (l / bl == r / bl) {
        shift(l, r);
      } else {
        int last = del(r);
        cnt[r / bl][last]--;
        add(l, last);
        cnt[l / bl][last]++;
        for (int i = l / bl; i <= r / bl; i++) {
          if (i > l / bl) {
            arr[i].push_front(last);
            cnt[i][last]++;
          }
          if (i < r / bl) {
            int first = arr[i].back();
            arr[i].pop_back();
            cnt[i][first]--;
            last = first;
          }
        }
      }
    } else {
      cin >> k;
      l = (l + lastans - 1) % n + 1;
      r = (r + lastans - 1) % n + 1;
      k = (k + lastans - 1) % n + 1;
      if (l > r) swap(l, r);
      int ans = 0;
      if (l / bl == r / bl) {
        int b = l / bl;
        for (int i = l - b * bl; i <= r - b * bl; i++) {
          if (arr[b][i] == k) ans++;
        }
      } else {
        int b = l / bl;
        for (int i = l; i < (b + 1) * bl; i++) {
          if (arr[b][i - b * bl] == k) ans++;
        }
        b = r / bl;
        for (int i = b * bl; i <= r; i++) {
          if (arr[b][i - b * bl] == k) ans++;
        }
        for (int i = l / bl + 1; i < r / bl; i++) {
          ans += cnt[i][k];
        }
      }
      cout << ans << "\n";
      lastans = ans;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  int T = 1;
  while (T--) {
    solve();
    cout << "\n";
  }
}