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 INF = 1 << 30;
const int MAXN = 100010;
const int SIZE = 350;
const int CNT = 310;
deque<int> seq[CNT];
int cnt[CNT][MAXN];
int N;
int query(int a, int b, int k) {
  int ans = 0;
  int start = a / SIZE;
  int end = b / SIZE;
  deque<int>::iterator it1 = seq[start].begin() + a % SIZE;
  for (int i = a; i < (start + 1) * SIZE && i <= b; i++) {
    if (*it1 == k) ans++;
    ++it1;
  }
  if (start != end) {
    deque<int>::iterator it2 = seq[end].begin();
    for (int i = end * SIZE; i <= b; i++) {
      if (*it2 == k) ans++;
      ++it2;
    }
  }
  for (int i = start + 1; i < end; i++) ans += cnt[i][k];
  return ans;
}
void update(int a, int b) {
  int start = a / SIZE;
  int end = b / SIZE;
  int last = seq[end].at(b % SIZE);
  seq[end].erase(seq[end].begin() + b % SIZE);
  cnt[end][last]--;
  seq[start].insert(seq[start].begin() + a % SIZE, last);
  cnt[start][last]++;
  for (int i = start; i < end; i++) {
    int v = seq[i].back();
    seq[i].pop_back();
    cnt[i][v]--;
    seq[i + 1].push_front(v);
    cnt[i + 1][v]++;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> N;
  for (int i = 0; i < N; i++) {
    int x;
    cin >> x;
    seq[i / SIZE].push_back(x);
    cnt[i / SIZE][x]++;
  }
  int q;
  cin >> q;
  int prev = 0;
  while (q--) {
    int op;
    cin >> op;
    int l, r;
    cin >> l >> r;
    l = (l + prev - 1) % N;
    r = (r + prev - 1) % N;
    if (l > r) swap(l, r);
    if (op == 1) {
      update(l, r);
    } else {
      int k;
      cin >> k;
      k = (k + prev - 1) % N + 1;
      cout << (prev = query(l, r, k)) << endl;
    }
  }
  return 0;
}