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 long long B = 360;
const long long N = 100005;
long long block(long long x) { return x / B; }
long long offset(long long x) { return x % B; }
deque<long long> a[B];
signed c[B][N];
long long n, q;
signed main() {
  ios::sync_with_stdio(false);
  cin >> n;
  for (long long i = 0; i < n; i++) {
    long long x;
    cin >> x;
    a[block(i)].push_back(x);
    c[block(i)][x]++;
  }
  cin >> q;
  long long lastans = 0;
  for (long long i = 0; i < q; i++) {
    long long op, l, r, k;
    cin >> op >> l >> r;
    l = (l + lastans - 1) % n;
    r = (r + lastans - 1) % n;
    if (l > r) swap(l, r);
    if (op == 2) cin >> k;
    k = (k + lastans - 1) % n + 1;
    long long L = block(l);
    long long R = block(r);
    l = offset(l);
    r = offset(r);
    if (op == 1) {
      if (L == R) {
        long long x = a[R][r];
        a[R].erase(a[R].begin() + r);
        a[L].insert(a[L].begin() + l, x);
      } else {
        for (long long i = L; i < R; i++) {
          long long x = a[i].back();
          a[i].pop_back();
          c[i][x]--;
          c[i + 1][x]++;
          a[i + 1].push_front(x);
        }
        long long x = a[R][r + 1];
        a[R].erase(a[R].begin() + r + 1);
        a[L].insert(a[L].begin() + l, x);
        c[R][x]--;
        c[L][x]++;
      }
    } else {
      long long ans = 0;
      if (L == R) {
        for (long long i = l; i <= r; i++) ans += a[L][i] == k;
      } else {
        for (long long i = l; i < a[L].size(); i++) ans += a[L][i] == k;
        for (long long i = 0; i <= r; i++) ans += a[R][i] == k;
        for (long long i = L + 1; i <= R - 1; i++) ans += c[i][k];
      }
      cout << (lastans = ans) << endl;
    }
  }
}