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 M = 320;
void read(int &x) {
  char ch = getchar();
  x = 0;
  for (; ch == ' ' || ch == '\n'; ch = getchar())
    ;
  for (; ch != ' ' && ch != '\n'; x = x * 10 + ch - '0', ch = getchar())
    ;
}
void write(int x) {
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
}
int n, q;
deque<int> a[325];
int cnt[325][100004];
int l, r, k, ans, oper;
void go(int &x) { x = ((x + ans - 1) % n) + 1; }
int main() {
  read(n);
  for (int i = 0; i < n; i++) {
    int x;
    read(x);
    a[i / M].push_back(x);
    cnt[i / M][x]++;
  }
  read(q);
  while (q--) {
    read(oper), read(l), read(r);
    go(l), go(r);
    l--, r--;
    if (l > r) swap(l, r);
    if (oper == 2) {
      read(k), go(k);
      ans = 0;
      if (l / M == r / M) {
        for (int i = l % M; i <= r % M; i++)
          if (a[l / M][i] == k) ans++;
        write(ans), puts("");
        continue;
      }
      for (int i = l % M; i < a[l / M].size(); i++)
        if (a[l / M][i] == k) ans++;
      for (int i = l / M + 1; i < r / M; i++) ans += cnt[i][k];
      for (int i = 0; i <= r % M; i++)
        if (a[r / M][i] == k) ans++;
      write(ans), puts("");
      continue;
    }
    if (l / M == r / M) {
      int x = a[l / M][r % M];
      a[r / M].erase(a[r / M].begin() + r % M);
      a[r / M].insert(a[r / M].begin() + l % M, x);
      continue;
    }
    int x = a[r / M][r % M];
    a[r / M].erase(a[r / M].begin() + r % M), cnt[r / M][x]--;
    for (int i = l / M + 1; i <= r / M; i++)
      a[i].push_front(a[i - 1].back()), cnt[i][a[i - 1].back()]++,
          cnt[i - 1][a[i - 1].back()]--, a[i - 1].pop_back();
    a[l / M].insert(a[l / M].begin() + l % M, x), cnt[l / M][x]++;
  }
}