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, sq = 350;
int cnt[sq][maxn];
deque<int> dq[sq];
void shift(int l, int r) {
  int idl = l / sq, idr = r / sq;
  int dl = l % sq, dr = r % sq;
  int val = dq[idr][dr];
  dq[idr].erase(dq[idr].begin() + dr);
  dq[idl].insert(dq[idl].begin() + dl, val);
  cnt[idr][val]--;
  cnt[idl][val]++;
  for (int i = idl; i < idr; i++) {
    int val = dq[i].back();
    dq[i + 1].push_front(val);
    dq[i].pop_back();
    cnt[i][val]--;
    cnt[i + 1][val]++;
  }
}
int get(int l, int r, int x) {
  int idl = l / sq, idr = r / sq;
  int dl = l % sq, dr = r % sq;
  int res = 0;
  if (idl == idr) {
    for (int i = dl; i <= dr; i++) res += dq[idl][i] == x;
    return res;
  }
  for (int i = dl; i < dq[idl].size(); i++) res += dq[idl][i] == x;
  for (int i = 0; i <= dr; i++) res += dq[idr][i] == x;
  for (int i = idl + 1; i < idr; i++) res += cnt[i][x];
  return res;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0, u; i < n; i++)
    scanf("%d", &u), dq[i / sq].push_back(u), cnt[i / sq][u]++;
  int q, lastans = 0;
  scanf("%d", &q);
  while (q--) {
    int type, l, r;
    scanf("%d", &type), scanf("%d", &l), scanf("%d", &r);
    l = (l + lastans - 1) % n, r = (r + lastans - 1) % n;
    if (l > r) swap(l, r);
    if (type == 1)
      shift(l, r);
    else {
      int k;
      scanf("%d", &k);
      k = (k + lastans - 1) % n + 1;
      printf("%d\n", lastans = get(l, r, k));
    }
  }
  return 0;
}