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 = (int)1e5 + 7;
const int INF = (int)1e9 + 7;
const long long linf = (long long)1e18 + 1;
const int block = 350;
int cnt[block + 2][N];
deque<int> dq[block + 1];
void upd(int l, int r) {
  int L = l / block, R = r / block;
  int r_val = dq[R][r % block];
  cnt[R][r_val]--;
  dq[R].erase(dq[R].begin() + (r % block));
  cnt[L][r_val]++;
  dq[L].insert(dq[L].begin() + (l % block), r_val);
  for (int i = L; i < R; ++i) {
    int val = dq[i].back();
    --cnt[i][val];
    dq[i].pop_back();
    ++cnt[i + 1][val];
    dq[i + 1].push_front(val);
  }
}
int get(int l, int r, int k) {
  int L = l / block, R = r / block;
  int res = 0;
  if (L == R) {
    int p = 0;
    for (auto to : dq[L]) {
      if (l % block <= p && p <= r % block && to == k) ++res;
      ++p;
    }
  } else {
    int p = 0;
    for (auto to : dq[L]) {
      if (l % block <= p && to == k) ++res;
      ++p;
    }
    p = 0;
    for (auto to : dq[R]) {
      if (p <= r % block && to == k) ++res;
      ++p;
    }
    for (int i = L + 1; i <= R - 1; ++i) res += cnt[i][k];
  }
  return res;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    int x;
    scanf("%d", &x);
    dq[i / block].push_back(x);
    cnt[i / block][x]++;
  }
  int q;
  scanf("%d", &q);
  int lastans = 0;
  for (int i = 1; i <= q; ++i) {
    int type, l, r;
    scanf("%d %d %d", &type, &l, &r);
    l = ((l + lastans - 1) % n), r = ((r + lastans - 1) % n);
    if (l > r) swap(l, r);
    if (type == 2) {
      int k;
      scanf("%d", &k);
      k = ((k + lastans - 1) % n) + 1;
      int cur = get(l, r, k);
      printf("%d\n", cur);
      lastans = cur;
    } else
      upd(l, r);
  }
  return 0;
}