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;
deque<int>::iterator it;
int n, m, x, b, size, maxb, ans, tmp, a, aa, bb, v, k, l, r;
struct blo {
  deque<int> q;
  int num[100010];
} block[405];
void init() {
  size = pow(n, 0.618);
  if (n % size == 0)
    maxb = n / size;
  else
    maxb = n / size + 1;
  for (int i = 0; i < n; ++i) {
    b = i / size;
    scanf("%d", &x);
    block[b].q.push_back(x);
    block[b].num[x]++;
  }
}
void solve1() {
  if (a == b) {
    it = block[a].q.begin() + bb;
    tmp = *it;
    block[a].q.erase(block[a].q.begin() + bb);
    block[a].q.insert(block[a].q.begin() + aa, tmp);
  } else {
    it = block[b].q.begin() + bb;
    tmp = *it;
    block[b].num[tmp]--;
    block[b].q.erase(block[b].q.begin() + bb);
    for (int i = a; i < b; ++i) {
      x = block[i].q.back();
      block[i].q.pop_back();
      block[i].num[x]--;
      block[i + 1].num[x]++;
      block[i + 1].q.push_front(x);
    }
    block[a].num[tmp]++;
    block[a].q.insert(block[a].q.begin() + aa, tmp);
  }
}
void solve2() {
  scanf("%d", &v);
  v = (v + ans - 1) % n + 1;
  ans = 0;
  if (a == b) {
    for (it = block[a].q.begin() + aa; it <= block[a].q.begin() + bb; it++)
      if (*it == v) ans++;
  } else {
    for (int i = a + 1; i < b; ++i) ans += block[i].num[v];
    for (it = block[a].q.begin() + aa; it < block[a].q.end(); it++)
      if (*it == v) ans++;
    for (it = block[b].q.begin(); it <= block[b].q.begin() + bb; it++)
      if (*it == v) ans++;
  }
  printf("%d\n", ans);
}
int main() {
  scanf("%d", &n);
  ans = 0;
  init();
  scanf("%d", &m);
  for (int i = 1; i <= m; ++i) {
    scanf("%d%d%d", &k, &l, &r);
    l = (l + ans - 1) % n;
    r = (r + ans - 1) % n;
    if (l >= r) swap(l, r);
    a = l / size;
    b = r / size;
    aa = l % size;
    bb = r % size;
    if (k == 1)
      solve1();
    else
      solve2();
  }
  fclose(stdin);
  fclose(stdout);
  return 0;
}