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 sz = 320, N = 100005;
int cnt[sz][N];
deque<int> block[sz];
int qry(int l, int r, int k) {
  int lid = l / sz, rid = r / sz, L = l % sz, R = r % sz;
  int res = 0;
  if (lid == rid) {
    for (int i = L; i <= R; i++)
      if (block[lid][i] == k) res++;
    return res;
  }
  for (int i = L; i < sz; i++)
    if (block[lid][i] == k) res++;
  for (int i = lid + 1; i < rid; i++) res += cnt[i][k];
  for (int i = 0; i <= R; i++)
    if (block[rid][i] == k) res++;
  return res;
}
void upd(int l, int r) {
  if (l == r) return;
  int L = l % sz, R = r % sz, lid = l / sz, rid = r / sz;
  int sv = block[rid][R];
  if (lid == rid) {
    block[lid].erase(block[lid].begin() + R);
    block[lid].insert(block[lid].begin() + L, sv);
    return;
  }
  cnt[rid][sv]--;
  block[rid].erase(block[rid].begin() + R);
  for (int i = lid; i < rid; i++) {
    int x = block[i].back();
    block[i].pop_back();
    cnt[i][x]--;
    cnt[i + 1][x]++;
    block[i + 1].push_front(x);
  }
  block[lid].insert(block[lid].begin() + L, sv);
  cnt[lid][sv]++;
}
int main() {
  int n, q, t, lastans = 0, l, r, k;
  scanf("%d", &n);
  for (int i = 0, a; i < n; i++)
    scanf("%d", &a), cnt[i / sz][a]++, block[i / sz].push_back(a);
  scanf("%d", &q);
  while (q--) {
    scanf("%d%d%d", &t, &l, &r);
    l = ((l + lastans - 1) % n) + 1;
    r = ((r + lastans - 1) % n) + 1;
    if (l > r) swap(l, r);
    l--;
    r--;
    if (t == 1)
      upd(l, r);
    else {
      scanf("%d", &k);
      k = ((k + lastans - 1) % n) + 1;
      int res = qry(l, r, k);
      cout << res << "\n";
      lastans = res;
    }
  }
  return 0;
}