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;
deque<int> box[351];
int ile[351][100005];
int ans = 0, n, q, blocks;
inline void deal(int &x) { x = (x + ans - 1) % n + 1; }
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  blocks = sqrt(n);
  for (int i = 0; i < n; ++i) {
    int temp;
    cin >> temp;
    int block = i / blocks;
    box[block].push_back(temp);
    ile[block][temp]++;
  }
  cin >> q;
  while (q--) {
    int type, ll, rr, k;
    cin >> type >> ll >> rr;
    deal(ll);
    deal(rr);
    ll--;
    rr--;
    if (ll > rr) swap(ll, rr);
    if (type == 1) {
      int left = ll / blocks;
      int right = rr / blocks;
      if (left == right) {
        ll -= blocks * left;
        rr -= blocks * left;
        int value = box[left][rr];
        box[left].erase(box[left].begin() + rr);
        box[left].insert(box[left].begin() + ll, value);
      } else {
        ll -= blocks * left;
        rr -= blocks * right;
        int value = box[right][rr];
        box[left].insert(box[left].begin() + ll, value);
        ile[left][value]++;
        box[right].erase(box[right].begin() + rr);
        ile[right][value]--;
        for (int i = left; i < right; ++i) {
          int value = box[i].back();
          box[i].pop_back();
          ile[i][value]--;
          box[i + 1].push_front(value);
          ile[i + 1][value]++;
        }
      }
    } else {
      cin >> k;
      deal(k);
      ans = 0;
      int left = ll / blocks;
      int right = rr / blocks;
      ll -= blocks * left;
      rr -= blocks * right;
      if (left == right) {
        for (int i = ll; i <= rr; ++i) {
          if (box[left][i] == k) ans++;
        }
        cout << ans << endl;
      } else {
        for (int i = ll; i < box[left].size(); ++i) {
          if (box[left][i] == k) ans++;
        }
        for (int i = left + 1; i < right; ++i) {
          ans += ile[i][k];
        }
        for (int i = 0; i <= rr; ++i) {
          if (box[right][i] == k) ans++;
        }
        cout << ans << endl;
      }
    }
  }
}