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 N = 1e5 + 10, sz = 340;
int n, m, ans, l, r, t, k;
int f[sz][N];
deque<int> q[sz];
int arr[N];
void init() {
  for (int i = 0, id = 0; i < n; i += sz, id++) {
    for (int j = i; j < i + sz && j < n; ++j) {
      f[id][arr[j]]++;
      q[id].emplace_back(arr[j]);
    }
  }
}
void shift(int l, int r) {
  int t = q[r / sz][r % sz];
  f[r / sz][t]--;
  f[l / sz][t]++;
  q[r / sz].erase(q[r / sz].begin() + r % sz);
  int cur = t;
  for (int id = l / sz; id < r / sz; ++id) {
    t = q[id].back();
    q[id].pop_back();
    f[id][t]--;
    q[id + 1].emplace_front(t);
    f[id + 1][t]++;
  }
  q[l / sz].insert(q[l / sz].begin() + l % sz, cur);
}
int get(int l, int r, int k) {
  int ans = 0;
  while (l % sz && l < r) ans += q[l / sz][l % sz] == k, l++;
  while (l / sz < r / sz) ans += f[l / sz][k], l += sz;
  while (l <= r) ans += q[l / sz][l % sz] == k, l++;
  return ans;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) scanf("%d", arr + i);
  scanf("%d", &m);
  init();
  while (m--) {
    scanf("%d%d%d", &t, &l, &r);
    l = (ans + l - 1) % n;
    r = (ans + r - 1) % n;
    if (l > r) swap(l, r);
    if (t == 1)
      shift(l, r);
    else {
      scanf("%d", &k);
      k = (k + ans - 1) % n + 1;
      ans = get(l, r, k);
      printf("%d\n", ans);
    }
  }
  return 0;
}