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 = 100100, P = 350;
deque<int> d[P];
int num[P][N], lastans;
int findans(int l, int r, int k) {
  int ans = 0;
  while (l % P != 0 && l < r) {
    if (d[l / P][l % P] == k) ans++;
    l++;
  }
  r--;
  while (r % P != P - 1 && r > l) {
    if (d[r / P][r % P] == k) ans++;
    r--;
  }
  if (r == l && l % P == 0)
    if (d[l / P][l % P] == k) ans++;
  for (int i = l; i < r; i += P) ans += num[i / P][k];
  return ans;
}
void upd(int l, int r) {
  int il = l / P, ir = r / P, b = l % P, e = r % P;
  int t = d[ir][e];
  if (il == ir) {
    for (int i = e; i > b; i--) d[il][i] = d[il][i - 1];
    d[il][b] = t;
  } else {
    for (int i = il; i < ir; i++) {
      int tmp = d[i].back();
      d[i + 1].push_front(tmp);
      num[i + 1][tmp]++;
      num[i][tmp]--;
      d[i].pop_back();
    }
    d[il].push_back(0);
    for (int i = e + 1; i > 0; i--) d[ir][i] = d[ir][i - 1];
    for (int i = (int)d[il].size() - 1; i > b; i--) d[il][i] = d[il][i - 1];
    num[ir][t]--;
    num[il][t]++;
    d[ir].pop_front();
    d[il][b] = t;
  }
  return;
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int n, m, x, l, r, k, tp;
  cin >> n;
  for (int i = 0; i < n; i++) {
    int t = i / P;
    cin >> x;
    d[t].push_back(x);
    num[t][x]++;
  }
  cin >> m;
  for (int i = 0; i < m; i++) {
    cin >> tp >> l >> r;
    l = (l + lastans - 1) % n + 1, r = (r + lastans - 1) % n + 1;
    if (l > r) swap(l, r);
    l--;
    if (tp == 2) {
      cin >> k;
      k = (lastans + k - 1) % n + 1;
      lastans = findans(l, r, k);
      cout << lastans << endl;
    } else
      upd(l, r - 1);
  }
  return 0;
}