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 magic = 350;
const int maxn = 101000;
int a[maxn], cnt[magic][maxn];
deque<int> b[magic];
void modify(int l, int r) {
  --l;
  --r;
  int lg = l / magic, rg = r / magic;
  int ln = l - lg * magic, rn = r - rg * magic;
  int num = b[rg][rn];
  --cnt[rg][num];
  ++cnt[lg][num];
  b[rg].erase(b[rg].begin() + rn);
  b[lg].insert(b[lg].begin() + ln, num);
  for (int i = lg; i < rg; ++i) {
    num = b[i].back();
    b[i].pop_back();
    b[i + 1].push_front(num);
    --cnt[i][num];
    ++cnt[i + 1][num];
  }
}
int solve(int l, int r, int k) {
  --l;
  --r;
  int ret = 0;
  int lg = l / magic, rg = r / magic;
  for (int i = l - lg * magic, j = 0; i < magic && j < (r - l + 1); ++i, ++j) {
    if (b[lg][i] == k) ++ret;
  }
  if (lg == rg) return ret;
  for (int i = lg + 1; i < rg; ++i) ret += cnt[i][k];
  for (int i = 0; i <= r - rg * magic; ++i)
    if (b[rg][i] == k) ++ret;
  return ret;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    b[i / magic].push_back(a[i]);
    cnt[i / magic][a[i]]++;
  }
  int q, lastans = 0;
  cin >> q;
  while (q--) {
    int d, l, r;
    cin >> d >> l >> r;
    l = ((l + lastans - 1) % n) + 1;
    r = ((r + lastans - 1) % n) + 1;
    if (l > r) swap(l, r);
    if (d == 1) {
      modify(l, r);
    } else {
      int k;
      cin >> k;
      k = ((k + lastans - 1) % n) + 1;
      lastans = solve(l, r, k);
      cout << lastans << endl;
    }
  }
  return 0;
}