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;
int n, ar[100010], cnt[350][100010], ln;
deque<int> sq[350];
void shiftR(int L, int R) {
  int fb = L / ln, sb = R / ln;
  int fl = (L - (fb * ln)), sl = (R - (sb * ln));
  int val = sq[sb][sl];
  sq[sb].erase(sq[sb].begin() + sl);
  sq[fb].insert(sq[fb].begin() + fl, val);
  cnt[fb][val]++;
  cnt[sb][val]--;
  for (int i = fb; i < sb; i++) {
    int tmp = sq[i].back();
    sq[i].pop_back();
    sq[i + 1].push_front(tmp);
    cnt[i][tmp]--;
    cnt[i + 1][tmp]++;
  }
}
int query(int L, int R, int K) {
  int fb = L / ln, sb = R / ln, ret = 0;
  for (int i = L - (fb * ln), j = 0; i < ln && j < (R - L + 1); i++, j++) {
    if (sq[fb][i] == K) ret++;
  }
  if (fb == sb) return ret;
  for (int i = fb + 1; i < sb; i++) {
    ret += cnt[i][K];
  }
  for (int i = 0; i <= R - (sb * ln); i++) {
    if (sq[sb][i] == K) ret++;
  }
  return ret;
}
int main() {
  scanf("%d", &n);
  ln = ceil(sqrt(n));
  for (int i = 0; i < n; i++) {
    scanf("%d", ar + i);
    sq[i / ln].push_back(ar[i]);
    cnt[i / ln][ar[i]]++;
  }
  int q, l, r, typ, ans = 0, k;
  scanf("%d", &q);
  while (q--) {
    scanf("%d %d %d", &typ, &l, &r);
    l = ((l + ans - 1) % n) + 1;
    l--;
    r = ((r + ans - 1) % n) + 1;
    r--;
    if (r < l) swap(l, r);
    if (typ == 1) {
      shiftR(l, r);
    } else {
      scanf("%d", &k);
      k = ((k + ans - 1) % n) + 1;
      ans = query(l, r, k);
      printf("%d\n", ans);
    }
  }
}