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 MAX = 100005;
const int SQRT = 400;
deque<int> d[SQRT];
int cnt[SQRT][MAX], tmp[SQRT + 100], t;
void read(int x) {
  t = 0;
  for (deque<int>::iterator it = d[x].begin(); it != d[x].end(); it++)
    tmp[t++] = *it;
}
void recalc(int x) {
  while (!d[x].empty()) {
    cnt[x][d[x].back()]--;
    d[x].pop_back();
  }
  for (int i = 0; i < t; i++)
    if (tmp[i] != -1) {
      d[x].push_back(tmp[i]);
      cnt[x][tmp[i]]++;
    }
}
void rot(int l, int r) {
  if (l / SQRT == r / SQRT) {
    read(l / SQRT);
    rotate(tmp + l % SQRT, tmp + r % SQRT, tmp + r % SQRT + 1);
    recalc(l / SQRT);
  } else {
    for (int i = l / SQRT; i < r / SQRT; i++) {
      int val = d[i].back();
      d[i + 1].push_front(val);
      cnt[i + 1][val]++;
      d[i].pop_back();
      cnt[i][val]--;
    }
    read(r / SQRT);
    int last = tmp[r % SQRT + 1];
    tmp[r % SQRT + 1] = -1;
    recalc(r / SQRT);
    read(l / SQRT);
    for (int i = t; i > l % SQRT; i--) tmp[i] = tmp[i - 1];
    tmp[l % SQRT] = last;
    t++;
    recalc(l / SQRT);
  }
}
int get(int l, int r, int k) {
  int ans = 0;
  if (l / SQRT == r / SQRT) {
    read(l / SQRT);
    for (int i = l % SQRT; i < r % SQRT + 1; i++) ans += (tmp[i] == k);
  } else {
    read(l / SQRT);
    for (int i = l % SQRT; i < t; i++) ans += (tmp[i] == k);
    read(r / SQRT);
    for (int i = 0; i < r % SQRT + 1; i++) ans += (tmp[i] == k);
    for (int i = l / SQRT + 1; i < r / SQRT; i++) ans += cnt[i][k];
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    int val;
    cin >> val;
    d[i / SQRT].push_back(val);
    cnt[i / SQRT][val]++;
  }
  int q;
  cin >> q;
  int last_ans = 0;
  while (q--) {
    int t, l, r;
    cin >> t >> l >> r;
    l = (l + last_ans - 1) % n;
    r = (r + last_ans - 1) % n;
    if (l > r) swap(l, r);
    if (t == 1)
      rot(l, r);
    else {
      int k;
      cin >> k;
      k = (k + last_ans - 1) % n + 1;
      last_ans = get(l, r, k);
      cout << last_ans << endl;
    }
  }
  return 0;
}