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>
#pragma GCC optimize("Ofast")
using namespace std;
const int N = 1e5 + 100, sq = 350;
int t[sq][N];
deque<int> a[sq];
int n;
int lastans = 0;
int untihash(int x) { return ((x + lastans - 1) % n) + 1; }
void upd(int l, int r) {
  int pl = l / sq;
  int pr = r / sq;
  int w = a[r / sq][r % sq];
  t[r / sq][w]--;
  t[l / sq][w]++;
  a[r / sq].erase(a[pr].begin() + (r % sq));
  a[l / sq].insert(a[pl].begin() + (l % sq), w);
  for (int i = pl; i < pr; i++) {
    int w = a[i].back();
    t[i][w]--;
    t[i + 1][w]++;
    a[i + 1].push_front(w);
    a[i].pop_back();
  }
}
int get(int l, int r, int k) {
  int pl = l / sq;
  int pr = (r - 1) / sq;
  int ans = 0;
  while (l % sq != 0 && l < r) {
    if (a[l / sq][l % sq] == k) {
      ans++;
    }
    l++;
  }
  while (r % sq != 0 && l < r) {
    r--;
    if (a[r / sq][r % sq] == k) {
      ans++;
    }
  }
  for (int i = l / sq; i < r / sq; i++) {
    ans += t[i][k];
  }
  return ans;
}
int b[N];
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    t[i / sq][x]++;
    a[i / sq].push_back(x);
  }
  int q;
  cin >> q;
  for (int i = 0; i < q; i++) {
    int ti;
    cin >> ti;
    if (ti == 1) {
      int l, r;
      cin >> l >> r;
      l = untihash(l);
      r = untihash(r);
      if (l > r) {
        swap(l, r);
      }
      l--;
      r--;
      upd(l, r);
    } else {
      int l, r, k;
      cin >> l >> r >> k;
      l = untihash(l);
      r = untihash(r);
      if (l > r) {
        swap(l, r);
      }
      k = untihash(k);
      l--;
      r--;
      cout << get(l, r + 1, k) << '\n';
      lastans = get(l, r + 1, k);
    }
  }
}