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;
int n, a[100007], sz[350], bl[350][2 * 350], cnt[350][100007], nblock;
void insertbl(int num, int pos, long long val) {
  sz[num]++;
  for (int i = sz[num]; i > pos; --i) bl[num][i] = bl[num][i - 1];
  bl[num][pos] = val;
  cnt[num][val]++;
}
long long erasebl(int num, int pos) {
  long long val = bl[num][pos];
  sz[num]--;
  for (int i = pos; i <= sz[num]; ++i) bl[num][i] = bl[num][i + 1];
  cnt[num][val]--;
  return val;
}
void fromatoblock() {
  nblock = 1;
  for (int i = 1; i <= n; ++i) {
    insertbl(nblock, sz[nblock] + 1, a[i]);
    if (sz[nblock] >= 350) nblock++;
  }
}
void fromblocktoa() {
  long long curr = 0;
  for (int i = 1; i <= nblock; ++i) {
    for (int j = 1; j <= sz[i]; ++j) {
      long long val = bl[i][j];
      a[++curr] = val;
      cnt[i][val]--;
    }
    sz[i] = 0;
  }
}
pair<int, int> blpos(int pos) {
  for (int i = 1; i <= nblock; ++i) {
    if (pos <= sz[i]) return {i, pos};
    pos -= sz[i];
  }
}
long long query(int l, int r, long long k) {
  long long res = 0;
  pair<int, int> p1 = blpos(l), p2 = blpos(r);
  if (p1.first == p2.first) {
    for (int i = p1.second; i <= p2.second; ++i)
      if (bl[p1.first][i] == k) res++;
    return res;
  }
  for (int i = p1.first + 1; i <= p2.first - 1; ++i) res += cnt[i][k];
  for (int i = p1.second; i <= sz[p1.first]; ++i)
    if (bl[p1.first][i] == k) res++;
  for (int i = p2.second; i >= 1; --i)
    if (bl[p2.first][i] == k) res++;
  return res;
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  long long q;
  long long lastans = 0;
  cin >> q;
  fromatoblock();
  for (int i = 1; i <= q; ++i) {
    int op, l, r;
    cin >> op >> l >> r;
    l = (l + lastans - 1) % n + 1;
    r = (r + lastans - 1) % n + 1;
    if (l > r) swap(l, r);
    if (op == 1) {
      pair<int, int> p1 = blpos(l), p2 = blpos(r);
      long long val = erasebl(p2.first, p2.second);
      insertbl(p1.first, p1.second, val);
    } else {
      long long k;
      cin >> k;
      k = (k + lastans - 1) % n + 1;
      lastans = query(l, r, k);
      cout << lastans << endl;
    }
    if (i % 350 == 0) {
      fromblocktoa();
      fromatoblock();
    }
  }
  return 0;
}