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 = 100010;
const int M = 333;
int n, a[N], c[M][N];
deque<int> Q[M];
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", a + i);
  for (int i = 0; i < n; i++) Q[i / M].push_back(a[i]);
  for (int i = 0; i < n; i++) c[i / M][a[i]]++;
  int q, t, l, r, k, last = 0, L, R, tmp, ans;
  scanf("%d", &q);
  while (q--) {
    scanf("%d%d%d", &t, &l, &r);
    l = (l + last - 1) % n, r = (r + last - 1) % n;
    if (l > r) swap(l, r);
    if (t == 1) {
      L = l / M, R = r / M;
      for (int i = R - 1; i >= L; i--) {
        tmp = Q[i][M - 1];
        Q[i].pop_back();
        c[i][tmp]--, c[i + 1][tmp]++;
        Q[i + 1].push_front(tmp);
      }
      if (l == r) continue;
      if (L == R) {
        tmp = Q[R][r % M];
        Q[R].erase(Q[R].begin() + (r % M));
        Q[L].insert(Q[L].begin() + (l % M), tmp);
      } else {
        tmp = Q[R][r % M + 1];
        Q[R].erase(Q[R].begin() + r % M + 1);
        c[R][tmp]--, c[L][tmp]++;
        Q[L].insert(Q[L].begin() + (l % M), tmp);
      }
    } else {
      scanf("%d", &k);
      k = (k + last - 1) % n + 1;
      ans = 0;
      L = l / M, R = r / M;
      for (int i = L + 1; i < R; i++) ans += c[i][k];
      if (L == R) {
        for (int i = l; i <= r; i++) ans += Q[L][i % M] == k;
      } else {
        for (int i = l; i < L * M + M; i++) ans += Q[L][i % M] == k;
        for (int i = R * M; i <= r; i++) ans += Q[R][i % M] == k;
      }
      printf("%d\n", ans);
      last = ans;
    }
  }
  return 0;
}