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 MAXN = 100001;
const int SQRT = 317;
int n, a[MAXN], cnt[SQRT][MAXN];
deque<int> deq[SQRT];
int decode(int x, int y) { return ((x + y - 1) % n) + 1; }
int ask(int l, int r, int k) {
  int ret = 0;
  while (l < r && l % SQRT) {
    if (deq[l / SQRT][l % SQRT] == k) ret++;
    l++;
  }
  while (l < r && r % SQRT) {
    r--;
    if (deq[r / SQRT][r % SQRT] == k) ret++;
  }
  for (; l < r; l += SQRT) ret += cnt[l / SQRT][k];
  return ret;
}
void shift(int l, int r) {
  r--;
  if (l / SQRT == r / SQRT) {
    int gro = l / SQRT;
    int x = deq[gro][r % SQRT];
    for (int i = r; i >= l + 1; --i)
      deq[gro][i % SQRT] = deq[gro][(i - 1) % SQRT];
    deq[gro][l % SQRT] = x;
  } else {
    int lg = l / SQRT, rg = r / SQRT;
    int x = deq[rg][r % SQRT];
    deq[rg].erase(deq[rg].begin() + r % SQRT);
    deq[lg].insert(deq[lg].begin() + l % SQRT, x);
    cnt[rg][x]--;
    cnt[lg][x]++;
    for (int i = lg + 1; i < rg + 1; ++i) {
      deq[i].push_front(deq[i - 1].back());
      cnt[i][deq[i - 1].back()]++;
      cnt[i - 1][deq[i - 1].back()]--;
      deq[i - 1].pop_back();
    }
  }
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    deq[i / SQRT].push_back(a[i]);
    cnt[i / SQRT][a[i]]++;
  }
  int q;
  cin >> q;
  int ans = 0;
  for (int p = 0; p < q; ++p) {
    int t;
    cin >> t;
    int l, r;
    cin >> l >> r;
    l = decode(l, ans);
    r = decode(r, ans);
    if (l > r) swap(l, r);
    l--;
    if (t == 2) {
      int k;
      cin >> k;
      k = decode(k, ans);
      ans = ask(l, r, k);
      cout << ans << "\n";
    } else
      shift(l, r);
  }
  return 0;
}