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 = 1e5 + 5;
const int B = 2000;
int n, m, a[N], bl[N], bll[N], blr[N], cnt[N / B + 5][N];
list<int> lst[N / B + 5];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", a + i);
  for (int i = 1, l = 1, r; l <= n; l = r + 1, ++i) {
    r = min(n, l + B - 1);
    bll[i] = l;
    blr[i] = r;
    for (int j = l; j <= r; ++j) {
      bl[j] = i;
      ++cnt[i][a[j]];
      lst[i].push_back(a[j]);
    }
  }
  scanf("%d", &m);
  int ans = 0;
  while (m--) {
    int type, l, r;
    scanf("%d%d%d", &type, &l, &r);
    l = (l + ans - 1) % n + 1;
    r = (r + ans - 1) % n + 1;
    if (l > r) swap(l, r);
    if (type == 1) {
      auto it = lst[bl[r]].begin();
      for (int i = bll[bl[r]]; i < r; ++i) ++it;
      int tmp = *it;
      --cnt[bl[r]][tmp];
      ++cnt[bl[l]][tmp];
      lst[bl[r]].erase(it);
      it = lst[bl[l]].begin();
      for (int i = bll[bl[l]]; i < l; ++i) ++it;
      lst[bl[l]].insert(it, tmp);
      for (int i = bl[l]; i < bl[r]; ++i) {
        int t = lst[i].back();
        lst[i].pop_back();
        lst[i + 1].push_front(t);
        --cnt[i][t];
        ++cnt[i + 1][t];
      }
    } else {
      int k;
      scanf("%d", &k);
      k = (k + ans - 1) % n + 1;
      ans = 0;
      if (bl[l] == bl[r]) {
        auto it = lst[bl[l]].begin();
        for (int i = bll[bl[l]]; i <= r; ++i) {
          if (i >= l) ans += *it == k;
          ++it;
        }
      } else {
        auto it = lst[bl[l]].begin();
        for (int i = bll[bl[l]]; i <= blr[bl[l]]; ++i) {
          if (i >= l) ans += *it == k;
          ++it;
        }
        it = lst[bl[r]].begin();
        for (int i = bll[bl[r]]; i <= r; ++i) {
          ans += *it == k;
          ++it;
        }
        for (int i = bl[l] + 1; i < bl[r]; ++i) ans += cnt[i][k];
      }
      printf("%d\n", ans);
    }
  }
  return 0;
}