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
#include <bits/stdc++.h>
using namespace std;
int blk, X, Y, x, y, A, L, R, k, i, n, c[350][100005], ans, j, Q, m;
deque<int> a[350];
int work(int x) { return (x - 1) / blk; }
int main() {
  blk = 300;
  scanf("%d", &n);
  for (i = 1; i <= n; i++)
    scanf("%d", &A), X = work(i), a[X].push_back(A), c[X][A]++;
  scanf("%d", &m);
  for (i = 1; i <= m; i++) {
    scanf("%d%d%d", &A, &L, &R);
    L = (L + ans - 1) % n;
    R = (R + ans - 1) % n;
    if (L > R) swap(L, R);
    L++;
    R++;
    X = work(L);
    Y = work(R);
    x = (L - 1) % blk;
    y = (R - 1) % blk;
    if (A == 1) {
      for (j = X + 1; j <= Y; j++) {
        c[j][a[j - 1].back()]++;
        a[j].push_front(a[j - 1].back());
      }
      for (j = X; j < Y; j++) {
        c[j][a[j].back()]--;
        a[j].pop_back();
      }
      if (X != Y)
        Q = a[Y][y + 1];
      else
        Q = a[Y][y];
      a[Y].erase(a[Y].begin() + y + (X != Y));
      c[Y][Q]--;
      a[X].insert(a[X].begin() + x, Q);
      c[X][Q]++;
    } else {
      scanf("%d", &k);
      k = (k + ans - 1) % n + 1;
      ans = 0;
      for (j = X + 1; j < Y; j++) ans += c[j][k];
      if (X != Y) {
        for (j = x; j < blk; j++) ans += (a[X][j] == k);
        for (j = 0; j <= y; j++) ans += (a[Y][j] == k);
      } else
        for (j = x; j <= y; j++) ans += (a[X][j] == k);
      printf("%d\n", ans);
    }
  }
}
