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;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cerr << name << " : " << arg1 << std::endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
const int N = int(1e5) + 10, SQRT = 320;
vector<int> arr(N);
int cnt[N / SQRT + 10][N], n, q;
deque<int> a[N / SQRT + 10];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  for (int i = (int)0; i < (int)n; i++) {
    cin >> arr[i];
    a[i / SQRT].push_back(arr[i]);
    cnt[i / SQRT][arr[i]]++;
  }
  cin >> q;
  int lastans = 0;
  for (int i = (int)0; i < (int)q; i++) {
    int choice;
    cin >> choice;
    if (choice == 1) {
      int L, R;
      cin >> L >> R;
      L = (L + lastans - 1 + n) % n;
      R = (R + lastans - 1 + n) % n;
      if (R < L) swap(R, L);
      int boxL = L / SQRT, posL = L % SQRT, boxR = R / SQRT, posR = R % SQRT;
      int x = a[boxR][posR];
      a[boxR].erase(a[boxR].begin() + posR);
      cnt[boxR][x]--;
      a[boxL].insert(a[boxL].begin() + posL, x);
      cnt[boxL][x]++;
      for (int box = boxL; box < boxR; box++) {
        int x = a[box].back();
        a[box].pop_back();
        cnt[box][x]--;
        a[box + 1].push_front(x);
        cnt[box + 1][x]++;
      }
    } else {
      int L, R, K;
      cin >> L >> R >> K;
      L = (L + lastans - 1 + n) % n;
      R = (R + lastans - 1 + n) % n;
      K = (K + lastans - 1 + n) % n + 1;
      if (R < L) swap(R, L);
      int boxL = L / SQRT, posL = L % SQRT, boxR = R / SQRT, posR = R % SQRT;
      int c = 0;
      if (boxL == boxR) {
        for (int pos = posL; pos <= posR; pos++) {
          c += a[boxL][pos] == K;
        }
      } else {
        for (int pos = posL; pos < SQRT; pos++) {
          c += a[boxL][pos] == K;
        }
        for (int pos = 0; pos <= posR; pos++) {
          c += a[boxR][pos] == K;
        }
        for (int box = boxL + 1; box < boxR; box++) {
          c += cnt[box][K];
        }
      }
      lastans = c;
      cout << lastans << endl;
    }
  }
  return 0;
}
