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 Maxn = 100005;
const int Maxm = 317;
int n;
int a[Maxn];
int q;
int res = 0;
int cnt[Maxm][Maxn];
deque<int> block[Maxm];
void Rot(int l, int r) {
  int el = block[r / Maxm][r % Maxm];
  block[r / Maxm].erase(block[r / Maxm].begin() + r % Maxm);
  cnt[r / Maxm][el]--;
  for (int i = r / Maxm - 1; i >= l / Maxm; i--) {
    block[i + 1].push_front(block[i].back());
    cnt[i + 1][block[i].back()]++;
    cnt[i][block[i].back()]--;
    block[i].pop_back();
  }
  block[l / Maxm].insert(block[l / Maxm].begin() + l % Maxm, el);
  cnt[l / Maxm][el]++;
}
int Calc(int l, int r, int k) {
  int res = 0;
  if (l / Maxm == r / Maxm) {
    for (int j = l % Maxm; j <= r % Maxm; j++) res += block[l / Maxm][j] == k;
  } else {
    for (int j = l % Maxm; j < block[l / Maxm].size(); j++)
      res += block[l / Maxm][j] == k;
    for (int i = l / Maxm + 1; i < r / Maxm; i++) res += cnt[i][k];
    for (int j = 0; j <= r % Maxm; j++) res += block[r / Maxm][j] == k;
  }
  return res;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    cnt[i / Maxm][a[i]]++;
    block[i / Maxm].push_back(a[i]);
  }
  scanf("%d", &q);
  while (q--) {
    int typ, l, r, k;
    scanf("%d %d %d", &typ, &l, &r);
    l = (l + res - 1 + n) % n;
    r = (r + res - 1 + n) % n;
    if (l > r) swap(l, r);
    if (typ == 1)
      Rot(l, r);
    else {
      scanf("%d", &k);
      k = (k + res - 1 + n) % n + 1;
      res = Calc(l, r, k);
      printf("%d\n", res);
    }
  }
  return 0;
}