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;
const int MAXB = 340;
deque<int> q[MAXB];
int BL, n, lasans, cnt[MAXB][MAXN], bl[MAXN], st[MAXB], en[MAXB];
int main() {
  cin >> n;
  BL = int(sqrt(n));
  for (int i = 0; i < n; i++) {
    bl[i] = i / BL;
    if (!st[bl[i]]) st[bl[i]] = i;
    en[bl[i]] = i;
    int x;
    cin >> x;
    cnt[bl[i]][x]++;
    q[bl[i]].push_back(x);
  }
  int Q;
  cin >> Q;
  while (Q--) {
    int op, l, r, x, y, k;
    cin >> op >> l >> r;
    l = (l + lasans - 1) % n + 1;
    r = (r + lasans - 1) % n + 1;
    if (l > r) swap(l, r);
    l--;
    x = l / BL;
    y = r / BL;
    if (op == 1) {
      if (x == y) {
        l %= BL;
        r = r % BL - 1;
        int num = q[x][r];
        q[x].erase(q[x].begin() + r);
        q[x].insert(q[x].begin() + l, num);
      } else {
        for (int i = x; i < y; i++) {
          int num = q[i].back();
          cnt[i][num]--;
          q[i].pop_back();
          q[i + 1].push_front(num);
          cnt[i + 1][num]++;
        }
        r %= BL;
        int num = q[y][r];
        q[y].erase(q[y].begin() + r);
        cnt[y][num]--;
        q[x].insert(q[x].begin() + l % BL, num);
        cnt[x][num]++;
      }
    } else {
      cin >> k;
      k = (k + lasans - 1) % n + 1;
      if (x == y) {
        lasans = 0;
        for (int i = l % BL; i < r % BL; i++) lasans += q[x][i] == k;
      } else {
        lasans = 0;
        for (int i = x + 1; i < y; i++) lasans += cnt[i][k];
        for (int i = l % BL; i < q[x].size(); i++) lasans += q[x][i] == k;
        for (int i = 0; i < r % BL; i++) lasans += q[y][i] == k;
      }
      cout << lasans << endl;
    }
  }
  return 0;
}