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 n, ans, t, l, r;
int rb, lb;
struct _bk : list<int> {
  int cnt[100001];
  void blockrotate(int bl, int br) {
    if (bl == br) return;
    auto it = next(begin(), bl);
    auto jt = next(it, br - bl);
    splice(it, *this, jt);
  }
  void blockcnt(int bl, int br) {
    auto it = next(begin(), bl);
    br -= bl;
    for (int i = 0; i <= br; ++i) {
      if (*it == t) ++ans;
      ++it;
    }
  }
} block[320];
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &t);
    block[i / 1200].push_back(t), ++block[i / 1200].cnt[t];
  }
  int ___L___;
  scanf("%d", &___L___);
  while (___L___--) {
    scanf("%d%d%d", &t, &l, &r);
    l = (l + ans - 1) % n;
    r = (r + ans - 1) % n;
    if (l > r) swap(l, r);
    lb = l / 1200;
    rb = r / 1200;
    if (t == 1) {
      if (lb == rb)
        block[lb].blockrotate(l % 1200, r % 1200);
      else {
        auto it = next(block[lb].begin(), l % 1200);
        auto jt = next(block[rb].begin(), r % 1200);
        --block[rb].cnt[*jt];
        ++block[lb].cnt[*jt];
        block[lb].insert(it, *jt);
        block[rb].erase(jt);
        for (int i = lb; i < rb; ++i) {
          auto &L = block[i], &R = block[i + 1];
          --L.cnt[L.back()];
          ++R.cnt[L.back()];
          R.splice(R.begin(), L, --L.end());
        }
      }
    } else {
      scanf("%d", &t);
      t = (t + ans - 1) % n + 1;
      ans = 0;
      if (lb == rb)
        block[lb].blockcnt(l % 1200, r % 1200);
      else {
        auto it = next(block[lb].begin(), l % 1200);
        for (; it != block[lb].end(); ++it)
          if (*it == t) ++ans;
        it = block[rb].begin();
        for (int i = 0; i <= r % 1200; ++i) {
          if (*it == t) ++ans;
          ++it;
        }
        for (int i = lb + 1; i < rb; ++i) ans += block[i].cnt[t];
      }
      printf("%d\n", ans);
    }
  }
}
