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
#include <bits/stdc++.h>
using namespace std;
struct bck {
  int sz, cnt[100010], a[650];
};
int N, Q, a[100010], sqr, K;
bck el[330];
void rebuild() {
  int i, j, t = 0;
  for (i = 1; i <= K; i++)
    for (j = 1; j <= el[i].sz; j++) {
      a[++t] = el[i].a[j];
      el[i].cnt[el[i].a[j]] = 0;
    }
  K = 0;
  for (i = 1; i <= N; i += sqr) {
    K++;
    el[K].sz = 0;
    for (j = i; j <= min(N, i + sqr - 1); j++) {
      el[K].sz++;
      el[K].a[el[K].sz] = a[j];
      el[K].cnt[a[j]]++;
    }
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> N;
  int i, j, l, r, k, ans = 0, op, tot, t, ind, val, ind2;
  sqr = 1;
  while (sqr * sqr <= N) sqr++;
  sqr--;
  for (i = 1; i <= N; i++) cin >> a[i];
  for (i = 1; i <= N; i += sqr) {
    K++;
    el[K].sz = 0;
    for (j = i; j <= min(N, i + sqr - 1); j++) {
      el[K].sz++;
      el[K].a[el[K].sz] = a[j];
      el[K].cnt[a[j]]++;
    }
  }
  cin >> Q;
  for (t = 1; t <= Q; t++) {
    if (t % sqr == 0) rebuild();
    cin >> op >> l >> r;
    l = (ans + l - 1) % N + 1;
    r = (ans + r - 1) % N + 1;
    if (l > r) swap(l, r);
    if (op == 1) {
      if (l == r) continue;
      tot = 0;
      for (ind = 1; tot + el[ind].sz < r; ind++) tot += el[ind].sz;
      val = el[ind].a[r - tot];
      for (i = r - tot; i < el[ind].sz; i++) el[ind].a[i] = el[ind].a[i + 1];
      el[ind].cnt[val]--;
      el[ind].sz--;
      tot = 0;
      for (ind = 1; tot + el[ind].sz < l; ind++) tot += el[ind].sz;
      for (i = el[ind].sz + 1; i > l - tot; i--)
        el[ind].a[i] = el[ind].a[i - 1];
      el[ind].a[l - tot] = val;
      el[ind].cnt[val]++;
      el[ind].sz++;
    } else {
      cin >> k;
      k = (ans + k - 1) % N + 1;
      ans = 0, tot = 0;
      for (ind = 1; tot + el[ind].sz < l; ind++) tot += el[ind].sz;
      for (i = l - tot; i <= min(el[ind].sz, r - tot); i++)
        if (el[ind].a[i] == k) ans++;
      tot = 0;
      for (ind2 = 1; tot + el[ind2].sz < r; ind2++) {
        tot += el[ind2].sz;
        if (ind2 > ind) ans += el[ind2].cnt[k];
      }
      if (ind2 != ind) {
        for (i = 1; i <= r - tot; i++)
          if (el[ind2].a[i] == k) ans++;
      }
      cout << ans << "\n";
    }
  }
  return 0;
}
