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 + 100;
const int sqr = 328;
deque<int> block[sqr];
int ans[sqr][MAXN];
int n;
int lastans;
int gen(int k) { return ((k + lastans - 1) % n) + 1; }
void update(int l, int r) {
  int vr = block[r / sqr][r % sqr];
  block[r / sqr].erase(block[r / sqr].begin() + r % sqr);
  ans[r / sqr][vr]--;
  ans[l / sqr][vr]++;
  block[l / sqr].insert(block[l / sqr].begin() + l % sqr, vr);
  for (int id = l / sqr; id < r / sqr; id++) {
    int k = block[id].back();
    ans[id][k]--;
    ans[id + 1][k]++;
    block[id].pop_back();
    block[id + 1].push_front(k);
  }
}
int get(int l, int r, int w) {
  int cnt = 0;
  for (; l % sqr && l < r; l++) {
    cnt += (block[l / sqr][l % sqr] == w);
  }
  for (; r % sqr && r > l;) {
    r--;
    cnt += (block[r / sqr][r % sqr] == w);
  }
  for (int id = l / sqr; id < r / sqr; id++) {
    cnt += ans[id][w];
  }
  return cnt;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  int c;
  for (int i = 0; i < n; i++) {
    cin >> c;
    block[i / sqr].push_back(c);
    ans[i / sqr][c]++;
  }
  int q;
  cin >> q;
  while (q--) {
    int command;
    cin >> command;
    if (command == 1) {
      int l, r;
      cin >> l >> r;
      l = gen(l);
      r = gen(r);
      l--;
      r--;
      if (l > r) {
        swap(l, r);
      }
      update(l, r);
    } else {
      int cc = 0;
      int l, r, w;
      cin >> l >> r >> w;
      l = gen(l);
      r = gen(r);
      w = gen(w);
      if (l > r) {
        swap(l, r);
      }
      l--;
      lastans = get(l, r, w);
      cout << lastans << '\n';
    }
  }
}