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;
struct block {
  int uMap[100010];
  deque<int> dq;
};
vector<block> blocks;
int len, n;
vector<int> arr;
void build() {
  for (int i = 0; i < n; i++) {
    int pos = i / len;
    blocks[pos].dq.push_back(arr[i]);
    blocks[pos].uMap[arr[i]]++;
  }
}
void operation(int l, int r) {
  int lBlock = l / len;
  int rBlock = r / len;
  int lPos = l % len;
  int rPos = r % len;
  int val = blocks[rBlock].dq[rPos];
  blocks[rBlock].dq.erase(blocks[rBlock].dq.begin() + rPos);
  blocks[lBlock].dq.insert(blocks[lBlock].dq.begin() + lPos, val);
  blocks[rBlock].uMap[val]--;
  blocks[lBlock].uMap[val]++;
  for (int i = lBlock; i < rBlock; i++) {
    val = blocks[i].dq.back();
    blocks[i].dq.pop_back();
    blocks[i + 1].dq.push_front(val);
    blocks[i].uMap[val]--;
    blocks[i + 1].uMap[val]++;
  }
}
int query(int l, int r, int k) {
  int ans = 0;
  for (int i = l; i <= r;) {
    if (i % len == 0 && (i + len - 1) <= r) {
      ans += blocks[i / len].uMap[k], i += len;
    } else {
      ans += (blocks[i / len].dq[i % len] == k);
      i++;
    }
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  arr.resize(n);
  for (int i = 0; i < n; i++) cin >> arr[i];
  len = ceil(sqrt(n));
  blocks.resize(ceil(sqrt(n)));
  build();
  int q;
  cin >> q;
  int lastans = 0;
  for (int i = 0; i < q; i++) {
    int type;
    cin >> type;
    if (type == 1) {
      int l, r;
      cin >> l >> r;
      l = (l + lastans - 1) % n;
      r = (r + lastans - 1) % n;
      if (l > r) swap(l, r);
      operation(l, r);
    } else {
      int l, r, k;
      cin >> l >> r >> k;
      l = (l + lastans - 1) % n;
      r = (r + lastans - 1) % n;
      k = (k + lastans - 1) % n + 1;
      if (l > r) swap(l, r);
      lastans = query(l, r, k);
      cout << lastans << "\n";
    }
  }
  return 0;
}