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 SQRT = 317;
const int MAXN = 100001;
int a[SQRT][SQRT];
int head[SQRT], front[SQRT];
int cnt[SQRT][MAXN];
int sz[SQRT];
void shift_block(int block, int l_pos, int r_pos) {
  int j = (r_pos + sz[block] - 1) % sz[block];
  for (int i = r_pos; i != l_pos; i = j, j = (j + sz[block] - 1) % sz[block]) {
    a[block][i] = a[block][j];
  }
}
void shift(int l, int r) {
  int l_bl = l / SQRT;
  int l_pos = (l % SQRT + head[l_bl]) % sz[l_bl];
  int r_bl = r / SQRT;
  int r_pos = (r % SQRT + head[r_bl]) % sz[r_bl];
  int push_val = a[r_bl][r_pos];
  if (l_bl == r_bl) {
    shift_block(l_bl, l_pos, r_pos);
    a[l_bl][l_pos] = push_val;
    return;
  }
  int last_val = a[l_bl][front[l_bl]];
  cnt[l_bl][last_val]--;
  shift_block(l_bl, l_pos, front[l_bl]);
  a[l_bl][l_pos] = push_val;
  cnt[l_bl][push_val]++;
  cnt[r_bl][push_val]--;
  push_val = last_val;
  for (int i = l_bl + 1; i < r_bl; i++) {
    last_val = a[i][front[i]];
    cnt[i][last_val]--;
    cnt[i][push_val]++;
    head[i] = front[i];
    front[i] = (front[i] + sz[i] - 1) % sz[i];
    a[i][head[i]] = push_val;
    push_val = last_val;
  }
  cnt[r_bl][push_val]++;
  shift_block(r_bl, head[r_bl], r_pos);
  a[r_bl][head[r_bl]] = push_val;
}
int get_block(int block, int l_pos, int r_pos, int k) {
  int res = 0;
  for (int i = r_pos;; i = (i + sz[block] - 1) % sz[block]) {
    if (a[block][i] == k) {
      res++;
    }
    if (i == l_pos) {
      break;
    }
  }
  return res;
}
int get(int l, int r, int k) {
  int l_bl = l / SQRT;
  int l_pos = (l % SQRT + head[l_bl]) % sz[l_bl];
  int r_bl = r / SQRT;
  int r_pos = (r % SQRT + head[r_bl]) % sz[r_bl];
  if (l_bl == r_bl) {
    return get_block(l_bl, l_pos, r_pos, k);
  }
  int res = 0;
  res += get_block(l_bl, l_pos, front[l_bl], k);
  for (int i = l_bl + 1; i < r_bl; i++) {
    res += cnt[i][k];
  }
  res += get_block(r_bl, head[r_bl], r_pos, k);
  return res;
}
void solve() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    int bl = i / SQRT;
    int pos = i % SQRT;
    front[bl] = pos;
    sz[bl]++;
    int val;
    scanf("%d", &val);
    a[bl][pos] = val;
    cnt[bl][val]++;
  }
  int q;
  scanf("%d", &q);
  int res = 0;
  for (int i = 0; i < q; i++) {
    int t, l, r, k;
    scanf("%d%d%d", &t, &l, &r);
    l = (l + res - 1) % n;
    r = (r + res - 1) % n;
    if (l > r) {
      swap(l, r);
    }
    if (t == 1) {
      shift(l, r);
    } else {
      scanf("%d", &k);
      k = (k + res - 1) % n + 1;
      res = get(l, r, k);
      printf("%d\n", res);
    }
  }
}
int main() { solve(); }