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 = 320;
int N, M;
short cnt[MAXN][SQRT];
deque<int> dq[SQRT];
int get(int x) { return dq[x / SQRT][x % SQRT]; }
int main() {
  if (fopen("input.txt", "r")) {
    freopen("input.txt", "r", stdin);
  }
  scanf("%d", &N);
  for (int i = 0, x; i < N; i++) {
    scanf("%d", &x);
    dq[i / SQRT].push_back(x);
    cnt[x][i / SQRT]++;
  }
  scanf("%d", &M);
  int lastans = 0;
  int qt, lt, rt, x;
  for (int qi = 0; qi < M; qi++) {
    scanf("%d %d %d", &qt, &lt, &rt);
    lt = (lt + lastans - 1) % N;
    rt = (rt + lastans - 1) % N;
    if (lt > rt) {
      swap(lt, rt);
    }
    if (qt == 1) {
      if (lt / SQRT == rt / SQRT) {
        deque<int> &d = dq[lt / SQRT];
        lt %= SQRT;
        rt %= SQRT;
        rotate(d.begin() + lt, d.begin() + rt, d.begin() + rt + 1);
      } else {
        int last = get(rt);
        dq[rt / SQRT].erase(dq[rt / SQRT].begin() + rt % SQRT);
        cnt[last][rt / SQRT]--;
        for (int i = rt / SQRT; i > lt / SQRT; i--) {
          int val = dq[i - 1].back();
          dq[i].push_front(val);
          cnt[val][i]++;
          dq[i - 1].pop_back();
          cnt[val][i - 1]--;
        }
        dq[lt / SQRT].insert(dq[lt / SQRT].begin() + lt % SQRT, last);
        cnt[last][lt / SQRT]++;
      }
    } else {
      scanf("%d", &x);
      x = (x + lastans - 1) % N + 1;
      int ans = 0;
      for (; lt <= rt && lt % SQRT; lt++) {
        ans += (get(lt) == x);
      }
      for (; rt >= lt && rt % SQRT != SQRT - 1; rt--) {
        ans += (get(rt) == x);
      }
      if (lt <= rt) {
        lt /= SQRT;
        rt /= SQRT;
        for (int i = lt; i <= rt; i++) {
          ans += cnt[x][i];
        }
      }
      lastans = ans;
      printf("%d\n", ans);
    }
  }
}