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
#include <bits/stdc++.h>
using namespace std;
const int NMAX = 335;
vector<vector<int> > B(NMAX);
int lastans, cnt[NMAX][100001];
int main() {
  int N;
  cin >> N;
  int len = (int)ceil(sqrt(N + .0));
  int num = 0, up = len;
  for (int i = 0; i < N; i++) {
    int a;
    cin >> a;
    if (i == up) {
      up += len;
      num++;
    }
    B[num].push_back(a);
    cnt[num][a]++;
  }
  int M;
  cin >> M;
  for (int t = 0; t < M; t++) {
    int type;
    scanf("%d", &type);
    if (type == 1) {
      int l, r;
      scanf("%d%d", &l, &r);
      l = (l + lastans - 1) % N, r = (r + lastans - 1) % N;
      if (l > r) swap(l, r);
      int sum = 0, cur = 0;
      for (int i = 0; i < NMAX; i++) {
        if (r >= sum && r < sum + (int)B[i].size()) {
          cur = B[i][r - sum];
          B[i].erase(B[i].begin() + (r - sum));
          cnt[i][cur]--;
          break;
        }
        sum += (int)B[i].size();
      }
      sum = 0;
      for (int i = 0; i < NMAX; i++) {
        if (l >= sum && l < sum + (int)B[i].size()) {
          B[i].insert(B[i].begin() + (l - sum), cur);
          cnt[i][cur]++;
          break;
        }
        sum += (int)B[i].size();
      }
    } else {
      int l, r, k;
      scanf("%d%d%d", &l, &r, &k);
      l = (l + lastans - 1) % N, r = (r + lastans - 1) % N,
      k = (k + lastans - 1) % N + 1;
      if (l > r) swap(l, r);
      int sum = 0, ans = 0;
      for (int i = 0; i < NMAX; i++) {
        if (sum >= l && sum + (int)B[i].size() - 1 <= r)
          ans += cnt[i][k];
        else if ((l >= sum && l < sum + (int)B[i].size()) ||
                 (r >= sum && r < sum + (int)B[i].size())) {
          for (int j = 0; j < (int)B[i].size(); j++) {
            if (sum + j >= l && sum + j <= r && B[i][j] == k) ans++;
          }
        }
        sum += (int)B[i].size();
      }
      lastans = ans;
      printf("%d\n", ans);
    }
    if (t % len == 0) {
      vector<int> list;
      for (int i = 0; i < NMAX; i++) {
        for (int j = 0; j < (int)B[i].size(); j++) {
          list.push_back(B[i][j]);
          cnt[i][B[i][j]]--;
        }
        B[i].clear();
      }
      num = 0, up = len;
      for (int i = 0; i < (int)list.size(); i++) {
        if (i == up) {
          num++;
          up += len;
        }
        B[num].push_back(list[i]);
        cnt[num][list[i]]++;
      }
    }
  }
  return 0;
}
