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;
template <class T, class U>
void ckmin(T &a, U b) {
  if (a > b) a = b;
}
template <class T, class U>
void ckmax(T &a, U b) {
  if (a < b) a = b;
}
int N, Q;
deque<int> block[363];
int freq[363][100013];
int ans;
void work(int l, int r) {
  int val = block[r / 363][r % 363];
  block[r / 363].erase(block[r / 363].begin() + (r % 363));
  freq[r / 363][val]--;
  block[l / 363].insert(block[l / 363].begin() + (l % 363), val);
  freq[l / 363][val]++;
  for (auto i = (l / 363); i < (r / 363); i++) {
    freq[i + 1][block[i].back()]++;
    freq[i][block[i].back()]--;
    block[i + 1].push_front(block[i].back());
    block[i].pop_back();
  }
}
int query(int l, int r, int k) {
  int res = 0;
  for (; l <= r && l % 363; l++) {
    res += (block[l / 363][l % 363] == k);
  }
  if (l > r) return res;
  for (; (r + 1) % 363; r--) {
    res += (block[r / 363][r % 363] == k);
  }
  r++;
  for (auto i = (l / 363); i < (r / 363); i++) {
    res += freq[i][k];
  }
  return res;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> N;
  for (auto i = (0); i < (N); i++) {
    int x;
    cin >> x;
    x--;
    block[i / 363].push_back(x);
    freq[i / 363][x]++;
  }
  cin >> Q;
  while (Q--) {
    int qid, l, r;
    cin >> qid >> l >> r;
    l = (l + ans - 1) % N;
    r = (r + ans - 1) % N;
    if (l > r) swap(l, r);
    if (qid == 1) {
      work(l, r);
    } else {
      int k;
      cin >> k;
      k = (k + ans - 1) % N;
      ans = query(l, r, k);
      cout << ans << '\n';
    }
  }
  return 0;
}