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 M = 1e9 + 7;
const int N = 1e5 + 5;
const int INF = 1e18;
const int K = 320;
int n, a[N], cnt[K][N], q, l, r, ty;
deque<int> d[K];
void rebuild() {
  for (int i = 0; i < n; i++) {
    cnt[i / K][a[i]]++;
    d[i / K].push_back(a[i]);
  }
}
void upd(int l, int r) {
  int posr = r % K, posl = l % K, bl = l / K, br = r / K;
  if (bl == br) {
    int z = d[bl][posr];
    d[bl].erase(d[bl].begin() + posr);
    d[bl].insert(d[bl].begin() + posl, z);
    return;
  }
  int z = d[br][posr];
  d[br].erase(d[br].begin() + posr);
  cnt[br][z]--;
  for (int i = br; i > bl; i--) {
    int x = d[i - 1].back();
    d[i].push_front(x);
    d[i - 1].pop_back();
    cnt[i][x]++;
    cnt[i - 1][x]--;
  }
  d[bl].insert(d[bl].begin() + posl, z);
  cnt[bl][z]++;
}
int get(int l, int r, int x) {
  int posr = r % K, posl = l % K, bl = l / K, br = r / K;
  int res = 0;
  if (bl == br) {
    for (int i = posl; i <= posr; i++) {
      if (d[bl][i] == x) res++;
    }
    return res;
  }
  for (int i = posl; i < d[bl].size(); i++) {
    if (d[bl][i] == x) res++;
  }
  for (int i = bl + 1; i < br; i++) {
    res += cnt[i][x];
  }
  for (int i = 0; i <= posr; i++) {
    if (d[br][i] == x) res++;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  rebuild();
  int last = 0;
  cin >> q;
  for (int test = 1; test <= q; test++) {
    cin >> ty >> l >> r;
    l = (l + last - 1) % n;
    r = (r + last - 1) % n;
    if (l > r) swap(l, r);
    if (ty == 1) {
      upd(l, r);
    } else {
      int k;
      cin >> k;
      k = (last + k - 1) % n + 1;
      last = get(l, r, k);
      cout << last << "\n";
    }
  }
  return 0;
}