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;
deque<int> a[355];
int n, m, S, Q, z, L, R, k, l, r, x, c[355][100005];
int ans;
void deal(int &x) { x = (x + ans - 1) % n + 1; }
int main() {
  scanf("%d", &n);
  S = int(sqrt(n)), m = n / S + 1;
  for (int i = 0; i < n; i++) {
    scanf("%d", &x);
    a[i / S].push_back(x);
    c[i / S][x]++;
  }
  scanf("%d", &Q);
  while (Q--) {
    scanf("%d%d%d", &z, &L, &R);
    deal(L), deal(R);
    if (L > R) swap(L, R);
    L--, l = L / S, r = R / S;
    if (z == 2) {
      scanf("%d", &k);
      deal(k);
      ans = 0;
      if (l == r) {
        for (int i = L % S, _ = R % S; i < _; i++) ans += a[l][i] == k;
      } else {
        for (int i = l + 1; i < r; i++) ans += c[i][k];
        for (int i = L % S, _ = a[l].size(); i < _; i++) ans += a[l][i] == k;
        for (int i = 0, _ = R % S; i < _; i++) ans += a[r][i] == k;
      }
      printf("%d\n", ans);
    } else {
      if (l == r) {
        R = R % S - 1, x = a[r][R], a[r].erase(a[r].begin() + R);
        a[l].insert(a[l].begin() + L % S, x);
      } else {
        for (int i = l; i < r;) {
          x = a[i].back(), a[i].pop_back(), c[i][x]--, i++;
          a[i].push_front(x), c[i][x]++;
        }
        R %= S, x = a[r][R], a[r].erase(a[r].begin() + R), c[r][x]--;
        a[l].insert(a[l].begin() + L % S, x), c[l][x]++;
      }
    }
  }
  return 0;
}