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 long long inf = 2e9 + 1e8;
const int N = 5e5 + 7;
int n, q, k, kol, a, f, l, r, ans, lans, y;
vector<vector<int> > x;
unordered_map<int, int> mp[330];
void build() {
  vector<int> lol;
  for (int i = 0; i <= k + 1; i++) {
    mp[i].clear();
    for (int j = 0; j < x[i].size(); j++) lol.push_back(x[i][j]);
    x[i].clear();
  }
  for (int i = 0; i < lol.size(); i++) {
    x[i / k].push_back(lol[i]);
    mp[i / k][lol[i]]++;
  }
  return;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  k = 320;
  x.resize(k + 5);
  for (int i = 0; i < n; i++) {
    cin >> a;
    x[0].push_back(a);
  }
  cin >> q;
  kol = int(sqrt(q));
  for (int j = 0; j < q; j++) {
    if (!(j % kol)) build();
    cin >> f >> l >> r;
    l--;
    r--;
    if (f == 1) {
      l = (l + lans) % n;
      r = (r + lans) % n;
      if (l > r) swap(l, r);
      int bl = 0, br = 0;
      while (x[bl].size() <= l) {
        l -= x[bl].size();
        bl++;
      }
      while (x[br].size() <= r) {
        r -= x[br].size();
        br++;
      }
      int lol = x[br][r];
      mp[bl][lol]++;
      mp[br][lol]--;
      x[br].erase(x[br].begin() + r);
      x[bl].insert(x[bl].begin() + l, lol);
    }
    if (f == 2) {
      cin >> y;
      l = (l + lans) % n;
      r = (r + lans) % n;
      y = (y + lans - 1) % n + 1;
      if (l > r) swap(l, r);
      int bl = 0, br = 0;
      while (x[bl].size() <= l) {
        l -= x[bl].size();
        bl++;
      }
      while (x[br].size() <= r) {
        r -= x[br].size();
        br++;
      }
      ans = 0;
      for (int b = bl, i = l; (b == br ? i <= r : b < br);)
        if (i == 0 && (b == br ? r == x[br].size() - 1 : true)) {
          if (mp[b].count(y)) ans += mp[b][y];
          b++;
        } else {
          ans += (x[b][i] == y);
          i++;
          if (i == x[b].size()) {
            i = 0;
            b++;
          }
        }
      lans = ans;
      cout << ans << '\n';
    }
  }
  return 0;
}