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;
vector<int> arr[1005];
vector<vector<int> > data((int)sqrt(1e5 + 2) + 5, vector<int>(100005, 0));
int sq, sz;
void reset() {
  for (int i = 0; i < sz; i++) {
    for (int j = 0; j < arr[i].size(); j++) {
      data[i][arr[i][j]] = 0;
    }
  }
  int ind = 0;
  vector<int> temp[sz];
  for (int i = 0; i < sz; i++) {
    for (int j = 0; j < arr[i].size(); j++) {
      temp[ind / sq].push_back(arr[i][j]);
      data[ind / sq][arr[i][j]]++;
      ind++;
    }
  }
  for (int i = 0; i < sz; i++) swap(temp[i], arr[i]);
}
pair<int, int> find(int x) {
  int ind = 0;
  for (int i = 0; i < sz; i++) {
    if (x - (int)arr[i].size() >= 0) {
      x -= arr[i].size();
    } else {
      ind = i;
      break;
    }
  }
  return {ind, x};
}
void erase(pair<int, int> x) {
  data[x.first][arr[x.first][x.second]]--;
  arr[x.first].erase(arr[x.first].begin() + x.second);
}
void add(pair<int, int> x, int v) {
  data[x.first][v]++;
  arr[x.first].insert(arr[x.first].begin() + x.second, v);
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  sq = sqrt(n);
  sz = (n - 1) / sq + 1;
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    arr[i / sq].push_back(x);
  }
  reset();
  int q;
  cin >> q;
  int lans = 0;
  while (q--) {
    int k;
    cin >> k;
    if (k == 1) {
      int x, y;
      cin >> x >> y;
      x = (x + lans - 1 + n) % n;
      y = (y + lans - 1 + n) % n;
      if (x > y) swap(x, y);
      pair<int, int> X = find(x);
      pair<int, int> Y = find(y);
      int v = arr[Y.first][Y.second];
      erase(Y);
      add(X, v);
      if (arr[X.first].size() >= 3 * sq) reset();
    } else {
      int x, y, v, ans = 0;
      cin >> x >> y >> v;
      x = (x + lans - 1 + n) % n;
      y = (y + lans - 1 + n) % n;
      v = (v + lans - 1 + n) % n + 1;
      if (x > y) swap(x, y);
      pair<int, int> X = find(x);
      pair<int, int> Y = find(y);
      int curr = X.first;
      int j = X.second;
      if (curr != Y.first && X.second != 0) {
        for (int i = X.second; i < arr[curr].size(); i++) {
          if (arr[curr][i] == v) ans++;
        }
        curr++;
        j = 0;
      }
      while (curr < Y.first) {
        ans += data[curr][v];
        curr++;
      }
      for (int i = j; i <= Y.second; i++) {
        if (arr[curr][i] == v) ans++;
      }
      cout << ans << "\n";
      lans = ans;
    }
  }
}