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>
#pragma comment(linker, "/STACK:36777216")
using namespace std;
const int N = 1e5 + 7;
const int INF = 0x3fffffff;
const int MOD = 1e9 + 7;
const long double EPS = 1e-8;
deque<int> Q[355];
int s[355][N];
int n, q, x, l, r, op, ans;
void gao(int &y) { y = (y + ans - 1) % n + 1; }
int main() {
  cin >> n;
  int Limit = sqrt(n + 0.5);
  for (int i = 0; i < n; i++) {
    scanf("%d", &x);
    Q[i / Limit].push_back(x);
    s[i / Limit][x]++;
  }
  cin >> q;
  while (q--) {
    scanf("%d", &op);
    if (op == 1) {
      scanf("%d%d", &l, &r);
      gao(l), gao(r);
      l--, r--;
      if (l > r) swap(l, r);
      int ll = l / Limit, rr = r / Limit;
      if (ll == rr) {
        int t = Q[rr][r % Limit];
        Q[rr].erase(Q[rr].begin() + (r % Limit));
        Q[ll].insert(Q[ll].begin() + (l % Limit), t);
      } else {
        for (int i = ll; i < rr; i++) {
          int t = Q[i].back();
          Q[i].pop_back();
          Q[i + 1].push_front(t);
          s[i][t]--, s[i + 1][t]++;
        }
        int t = Q[rr][1 + r % Limit];
        Q[rr].erase(Q[rr].begin() + (1 + r % Limit));
        s[rr][t]--;
        Q[ll].insert(Q[ll].begin() + (l % Limit), t);
        s[ll][t]++;
      }
    } else {
      scanf("%d%d%d", &l, &r, &x);
      gao(l), gao(r), gao(x);
      ans = 0;
      l--, r--;
      if (l > r) swap(l, r);
      int ll = l / Limit, rr = r / Limit;
      if (ll == rr) {
        for (int i = l % Limit, len = r % Limit; i <= len; i++) {
          ans += (Q[ll][i] == x);
        }
      } else {
        for (int i = ll + 1; i < rr; i++) ans += s[i][x];
        for (int i = l % Limit; i < Limit; i++) ans += (Q[ll][i] == x);
        for (int i = 0, len = r % Limit; i <= len; i++) ans += (Q[rr][i] == x);
      }
      printf("%d\n", ans);
    }
  }
}