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 long long MAX_N = 1e5 + 10, INF = 1e9, sq = 334;
int n, q, a[MAX_N], cnt[sq + 100][MAX_N], lastans;
deque<int> arr[sq + 100];
void upd(int l, int r) {
  int ind1 = l / sq, ind2 = r / sq;
  int x = arr[ind2][r % sq];
  if (ind1 != ind2) {
    for (int i = ind1; i < ind2; i++) {
      arr[i + 1].push_front(arr[i][arr[i].size() - 1]);
      cnt[i + 1][arr[i][arr[i].size() - 1]]++;
      cnt[i][arr[i][arr[i].size() - 1]]--;
      arr[i].pop_back();
    }
    arr[ind1].push_back(0);
    for (int i = arr[ind1].size() - 1; i > l % sq; i--) {
      arr[ind1][i] = arr[ind1][i - 1];
    }
    arr[ind1][l % sq] = x;
    cnt[ind1][x]++;
    cnt[ind2][x]--;
    for (int i = r % sq + 1; i < arr[ind2].size() - 1; i++) {
      arr[ind2][i] = arr[ind2][i + 1];
    }
    arr[ind2].pop_back();
  } else {
    int x = arr[ind1][r % sq];
    for (int i = r % sq; i > l % sq; i--) {
      arr[ind1][i] = arr[ind1][i - 1];
    }
    arr[ind1][l % sq] = x;
  }
}
int jav(int l, int r, int k) {
  int ret = 0;
  int ind1 = l / sq, ind2 = r / sq;
  if (ind1 != ind2) {
    for (int i = ind1 + 1; i < ind2; i++) {
      ret += cnt[i][k];
    }
    for (int i = l % sq; i < arr[ind1].size(); i++) {
      if (arr[ind1][i] == k) ret++;
    }
    for (int i = 0; i <= r % sq; i++) {
      if (arr[ind2][i] == k) ret++;
    }
  } else {
    for (int i = l % sq; i <= r % sq; i++) {
      if (arr[ind1][i] == k) ret++;
    }
  }
  return ret;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    arr[i / sq].push_back(a[i]);
    cnt[i / sq][a[i]]++;
  }
  cin >> q;
  while (q--) {
    int x, l, r, k;
    cin >> x;
    if (x == 1) {
      cin >> l >> r;
      l = ((l + lastans - 1) % n);
      r = ((r + lastans - 1) % n);
      if (l > r) swap(l, r);
      upd(l, r);
    } else {
      cin >> 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);
      lastans = jav(l, r, k);
      cout << lastans << endl;
    }
  }
}
