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;
long long int n, q, lastans, l, r, k, t, sq;
long long int a[100010];
int c[330][100010];
deque<long long int> p[330];
void query_1(long long int &a, long long int &b) {
  a = (a + lastans - 1) % n;
  b = (b + lastans - 1) % n;
  if (a > b) swap(a, b);
}
void query_2(long long int &a, long long int &b, long long int &key) {
  a = (a + lastans - 1) % n;
  b = (b + lastans - 1) % n;
  if (a > b) swap(a, b);
  key = (key + lastans - 1) % n + 1;
}
void cyclicswift() {
  query_1(l, r);
  long long int x = p[r / sq][r % sq];
  c[r / sq][x]--;
  c[l / sq][x]++;
  p[r / sq].erase(p[r / sq].begin() + r % sq);
  p[l / sq].insert(p[l / sq].begin() + l % sq, x);
  for (long long int i = l / sq; i < r / sq; i++) {
    x = p[i].back();
    p[i].pop_back();
    c[i][x]--;
    p[i + 1].push_front(x);
    c[i + 1][x]++;
  }
}
void countk() {
  query_2(l, r, k);
  long long int ans = 0;
  while (l <= r) {
    if ((l % sq == 0) && (l + sq - 1) <= r) {
      ans += c[l / sq][k];
      l = l + sq;
    } else {
      if (p[l / sq][l % sq] == k) ans++;
      l++;
    }
  }
  cout << ans << '\n';
  lastans = ans;
}
int main() {
  long long int i;
  cin >> n;
  sq = 317;
  for (i = 0; i < n; i++) {
    cin >> a[i];
    c[i / sq][a[i]]++;
    p[i / sq].push_back(a[i]);
  }
  lastans = 0;
  cin >> q;
  while (q--) {
    cin >> t;
    if (t == 1) {
      cin >> l >> r;
      cyclicswift();
    } else {
      cin >> l >> r >> k;
      countk();
    }
  }
  return 0;
}