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 = 1e5 + 10, SQRT = 340;
int a[MAXN], cnt[SQRT][MAXN];
deque<int> q[MAXN];
void build(int n) {
  for (int i = 0; i < n; i++) {
    q[i / SQRT].push_back(a[i]);
    cnt[i / SQRT][a[i]]++;
  }
}
void calc(int l, int r) {
  int a2 = q[r / SQRT][r % SQRT];
  cnt[r / SQRT][a2]--;
  q[r / SQRT].erase(q[r / SQRT].begin() + (r % SQRT));
  cnt[l / SQRT][a2]++;
  q[l / SQRT].insert(q[l / SQRT].begin() + (l % SQRT), a2);
  for (int i = (l / SQRT); i < (r / SQRT); i++) {
    int d = q[i].back();
    q[i].pop_back();
    cnt[i][d]--;
    q[i + 1].push_front(d);
    cnt[i + 1][d]++;
  }
}
int get(int l, int r, int val) {
  int ans = 0;
  bool is1 = false, is2 = false;
  int x1 = l / SQRT, x2 = r / SQRT;
  while (l % SQRT != 0 and l <= r)
    ans += (q[x1][l % SQRT] == val), l++, is1 = true;
  while ((r + 1) % SQRT != 0 and l <= r)
    ans += (q[x2][r % SQRT] == val), r--, is2 = true;
  if (is1) x1++;
  if (is2) x2--;
  for (int i = x1; i <= x2; i++) ans += cnt[i][val];
  return ans;
}
int main() {
  int n, m;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  build(n);
  cin >> m;
  int ans = 0;
  for (int i = 0; i < m; i++) {
    int x;
    cin >> x;
    if (x == 1) {
      int l, r;
      cin >> l >> r;
      l = (l + ans - 1 + n) % n + 1;
      r = (r + ans - 1 + n) % n + 1;
      if (l > r) swap(l, r);
      calc(l - 1, r - 1);
    } else {
      int l, r, val;
      cin >> l >> r >> val;
      l = (l + ans - 1 + n) % n + 1;
      r = (r + ans - 1 + n) % n + 1;
      if (l > r) swap(l, r);
      val = (val + ans - 1 + n) % n + 1;
      cout << get(l - 1, r - 1, val) << endl;
      ans = get(l - 1, r - 1, val);
    }
  }
}