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 long long INF = 1e9, MOD = INF + 7;
const int N = 1e5 + 5, SQ = 370;
int n, last, a[N], cnt[SQ][N];
deque<int> dq[SQ];
int decode(int x) { return (x + last - 1) % n + 1; }
void rebuild(int id) {
  int l = id * SQ, r = min(l + SQ, n);
  for (int i = l; i < r; i++) dq[id].push_back(a[i]), cnt[id][a[i]]++;
}
void shift(int l, int r) {
  r--;
  int x = dq[r / SQ][r % SQ];
  cnt[r / SQ][x]--;
  dq[r / SQ].erase(dq[r / SQ].begin() + r % SQ);
  cnt[l / SQ][x]++;
  dq[l / SQ].insert(dq[l / SQ].begin() + l % SQ, x);
  for (int id = l / SQ; id < r / SQ; id++) {
    x = dq[id].back();
    cnt[id][x]--;
    cnt[id + 1][x]++;
    dq[id].pop_back();
    dq[id + 1].push_front(x);
  }
}
int get(int l, int r, int k) {
  int ans = 0;
  while (l % SQ != 0 && l < r) ans += (dq[l / SQ][l % SQ] == k), l++;
  while (r % SQ != 0 && l < r) r--, ans += (dq[r / SQ][r % SQ] == k);
  for (int id = l / SQ; id < r / SQ; id++) ans += cnt[id][k];
  return ans;
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n; i += SQ) rebuild(i / SQ);
  int q;
  cin >> q;
  while (q--) {
    int t;
    cin >> t;
    int l, r;
    cin >> l >> r;
    l = decode(l), r = decode(r);
    if (l > r) swap(l, r);
    l--;
    if (t == 1)
      shift(l, r);
    else {
      int k;
      cin >> k;
      k = decode(k);
      last = get(l, r, k);
      cout << last << "\n";
    }
  }
  return 0;
}