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 MN = 1e5;
const int BS = 320;
int freq[BS][MN + 1];
deque<int> bucket[BS];
int shift(int k, int l, int r, int in) {
  int out = bucket[k][r];
  freq[k][out]--;
  for (int i = r; i > l; i--) {
    bucket[k][i] = bucket[k][i - 1];
  }
  bucket[k][l] = in;
  freq[k][in]++;
  return out;
}
int pos(int k, int i) { return i - k * BS; }
void update(int l, int r) {
  int i = l / BS, j = r / BS;
  if (i == j) {
    shift(i, pos(i, l), pos(j, r), bucket[i][pos(j, r)]);
  } else {
    int in = shift(i, pos(i, l), bucket[i].size() - 1, bucket[j][pos(j, r)]);
    for (int k = i + 1; k < j; k++) {
      bucket[k].push_front(in);
      freq[k][in]++;
      in = bucket[k].back();
      bucket[k].pop_back();
      freq[k][in]--;
    }
    shift(j, 0, pos(j, r), in);
  }
}
int query(int l, int r, int x) {
  int cnt = 0;
  int i = l / BS, j = r / BS;
  if (i == j) {
    for (int k = pos(i, l); k <= pos(j, r); k++) {
      cnt += (bucket[i][k] == x);
    }
  } else {
    for (int k = pos(i, l); k < bucket[i].size(); k++) {
      cnt += (bucket[i][k] == x);
    }
    for (int k = i + 1; k < j; k++) {
      cnt += freq[k][x];
    }
    for (int k = 0; k <= pos(j, r); k++) {
      cnt += (bucket[j][k] == x);
    }
  }
  return cnt;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    int x;
    scanf("%d", &x);
    int k = i / BS;
    bucket[k].push_back(x);
    freq[k][x]++;
  }
  int m;
  scanf("%d", &m);
  int last = 0;
  while (m--) {
    int o;
    scanf("%d", &o);
    switch (o) {
      case 1: {
        int l, r;
        scanf("%d %d", &l, &r);
        l = (l + last - 1) % n;
        r = (r + last - 1) % n;
        if (l > r) swap(l, r);
        update(l, r);
        break;
      }
      case 2: {
        int l, r, x;
        scanf("%d %d %d", &l, &r, &x);
        l = (l + last - 1) % n;
        r = (r + last - 1) % n;
        if (l > r) swap(l, r);
        x = (x + last - 1) % n + 1;
        last = (x > n ? 0 : query(l, r, x));
        printf("%d\n", last);
        break;
      }
    }
  }
}