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 BL = 205, BS = 500;
const int MAXN = 1e5 + 10;
int last = 0, n;
int cnt[BL][MAXN];
vector<int> blocks[BL], v;
int get(int p) {
  for (int i = 0; i < BL; i++) {
    if (p < blocks[i].size())
      return blocks[i][p];
    else
      p -= blocks[i].size();
  }
}
void rem(int p) {
  for (int i = 0; i < BL; i++) {
    if (p < blocks[i].size()) {
      cnt[i][blocks[i][p]]--;
      blocks[i].erase(blocks[i].begin() + p);
      return;
    } else
      p -= blocks[i].size();
  }
}
void add(int p, int va) {
  for (int i = 0; i < BL; i++) {
    if (p < blocks[i].size()) {
      cnt[i][va]++;
      blocks[i].insert(blocks[i].begin() + p, va);
      return;
    } else
      p -= blocks[i].size();
  }
}
void rebuild() {
  for (int i = 0; i < BL; i++) {
    for (auto x : blocks[i]) cnt[i][x]--;
    blocks[i].clear();
  }
  for (int i = 0; i < n; i++) {
    blocks[i / BS].push_back(v[i]);
    cnt[i / BS][v[i]]++;
  }
}
void print() {
  for (int i = 0; i < BL; i++)
    for (auto x : blocks[i]) cout << x << " ";
  cout << endl;
}
int qry(int l, int r, int k) {
  int st = 0, ans = 0;
  for (int i = 0; i < BL; i++) {
    if (st > r) break;
    if (st + blocks[i].size() - 1 < l) {
      st += blocks[i].size();
      continue;
    }
    if (l <= st && st + blocks[i].size() - 1 <= r)
      ans += cnt[i][k];
    else {
      for (int j = 0; j < blocks[i].size(); j++) {
        if (st + j >= l && st + j <= r && blocks[i][j] == k) {
          ans++;
        }
      }
    }
    st += blocks[i].size();
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  v.resize(n);
  for (auto &x : v) cin >> x;
  rebuild();
  int q, prev = 0;
  cin >> q;
  while (q--) {
    int t, l1, r1, k1 = 0;
    cin >> t >> l1 >> r1;
    if (t == 2) cin >> k1;
    int l = (l1 + last - 1) % n + 1;
    int r = (r1 + last - 1) % n + 1;
    int k = (k1 + last - 1) % n + 1;
    if (l > r) swap(l, r);
    if (t == 2) {
      last = qry(l - 1, r - 1, k);
      cout << last << '\n';
    } else {
      int x = get(r - 1);
      rem(r - 1);
      add(l - 1, x);
      prev++;
    }
    if (prev == 300) {
      prev = 0;
      v.clear();
      for (int i = 0; i < BL; i++)
        for (auto x : blocks[i]) v.push_back(x);
      rebuild();
    }
  }
}