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 long long MAXN = 1e5 + 100, mod = 1e9 + 7, inf = 1e17;
int n, m, x1, x2, d, z, t, sum, y, maxi, mini, ans, ptr, x, dp[330][MAXN], sq;
deque<int> q[MAXN];
void upd(long long x, long long y) {
  if (x / sq == y / sq) {
    for (int i = (y - 1) % sq; i >= (x % sq); i--) {
      swap(q[x / sq][i], q[x / sq][i + 1]);
    }
  } else {
    q[x / sq].insert(q[x / sq].begin() + (x % sq), q[y / sq][y % sq]);
    dp[x / sq][q[y / sq][y % sq]]++;
    dp[y / sq][q[y / sq][y % sq]]--;
    q[y / sq].erase(q[y / sq].begin() + (y % sq));
    for (int i = (x / sq) + 1; i <= y / sq; i++) {
      q[i].push_front(q[i - 1].back());
      dp[i][q[i - 1].back()]++;
      dp[i - 1][q[i - 1].back()]--;
      q[i - 1].pop_back();
    }
  }
}
long long query(long long x, long long y, long long z) {
  ans = 0;
  while (x <= y && x % sq != 0) {
    if (q[(x / sq)][x % sq] == z) ans++;
    x++;
  }
  while (x + sq <= y) {
    ans += dp[x / sq][z];
    x += sq;
  }
  while (x <= y) {
    if (q[(x / sq)][x % sq] == z) ans++;
    x++;
  }
  return ans;
}
int main() {
  cin >> n;
  sq = ceil(sqrt(n));
  for (int i = 0; i < n; i++) {
    cin >> x;
    q[i / sq].push_back(x);
    dp[i / sq][x]++;
  }
  cin >> m;
  long long lastans = 0;
  while (m--) {
    cin >> t;
    if (t == 1) {
      cin >> x >> y;
      x = ((x + lastans - 1) % n);
      y = ((y + lastans - 1) % n);
      if (x > y) {
        swap(x, y);
      }
      upd(x, y);
    } else {
      cin >> x >> y >> z;
      x = ((x + lastans - 1) % n);
      y = ((y + lastans - 1) % n);
      z = ((z + lastans - 1) % n) + 1;
      if (x > y) {
        swap(x, y);
      }
      lastans = query(x, y, z);
      cout << lastans << endl;
    }
  }
}