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 = 1e5 + 5;
deque<int> q[340];
int x[340][N];
int n, T, ans, cnt;
void decode(int &num) { num = (num + ans + n - 1) % n + 1; }
void Move(int l, int r) {
  r--;
  l--;
  int L = (l) / cnt, R = (r) / cnt;
  if (L == R) {
    int rem = q[R][r % cnt];
    q[R].erase(q[R].begin() + r % cnt);
    q[L].insert(q[L].begin() + l % cnt, rem);
    return;
  }
  q[L].insert(q[L].begin() + l % cnt, q[R][r % cnt]);
  x[L][q[R][r % cnt]]++;
  x[R][q[R][r % cnt]]--;
  q[R].erase(q[R].begin() + r % cnt);
  for (int i = L; i < R; i++) {
    x[i][q[i].back()]--;
    x[i + 1][q[i].back()]++;
    q[i + 1].push_front(q[i].back());
    q[i].pop_back();
  }
  return;
}
void search(int l, int r, int k) {
  l--, r--;
  int L = l / cnt, R = r / cnt;
  if (L == R)
    for (int i = l % cnt; i <= r % cnt; i++) ans += (q[L][i] == k);
  else {
    for (int i = l % cnt; i < q[L].size(); i++) ans += (q[L][i] == k);
    for (int i = L + 1; i < R; i++) ans += x[i][k];
    for (int i = 0; i <= r % cnt; i++) ans += (q[R][i] == k);
  }
  return;
}
int main() {
  cin >> n;
  cnt = (int)(sqrt(n));
  for (int i = 0; i < n; i++) {
    int a;
    cin >> a;
    x[i / cnt][a]++;
    q[i / cnt].push_back(a);
  }
  cin >> T;
  while (T--) {
    int l, r, Num;
    cin >> Num >> l >> r;
    decode(l);
    decode(r);
    if (l > r) swap(l, r);
    if (Num == 1) {
      Move(l, r);
    } else {
      int k;
      cin >> k;
      decode(k);
      ans = 0;
      search(l, r, k);
      cout << ans << endl;
    }
  }
  return 0;
}