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;
template <typename T>
inline T abs(T t) {
  return t < 0 ? -t : t;
}
const long long modn = 1000000007;
inline long long mod(long long x) { return x % modn; }
const int MAXN = 100010, SQRT = 320;
vector<deque<int> > bck;
int cnt[SQRT][MAXN];
int s[MAXN];
int n, m;
int main() {
  scanf("%d", &n);
  int szb = (int)ceil(sqrt(n));
  deque<int> aux;
  for (int a = 0; a < n; a++) {
    scanf("%d", &s[a]);
    if (a > 0 && (a - 1) / szb != a / szb) {
      bck.push_back(aux);
      aux.clear();
    }
    aux.push_back(s[a]);
    cnt[a / szb][s[a]]++;
  }
  if (!aux.empty()) bck.push_back(aux);
  scanf("%d", &m);
  int last_ans = 0;
  for (int a = 0; a < m; a++) {
    int t, i, j, k;
    scanf("%d%d%d", &t, &i, &j);
    i = (i + last_ans - 1) % n;
    j = (j + last_ans - 1) % n;
    if (i > j) swap(i, j);
    if (t == 1) {
      int fb = i / szb, lb = j / szb;
      int fs = i % szb, ls = j % szb;
      ;
      for (int a = ls; a < bck[lb].size() - 1; a++)
        swap(bck[lb][a], bck[lb][a + 1]);
      int ult = bck[lb].back();
      bck[lb].pop_back();
      ;
      cnt[lb][ult]--;
      bck[fb].push_front(ult);
      for (int a = 0; a < fs; a++) swap(bck[fb][a], bck[fb][a + 1]);
      cnt[fb][ult]++;
      for (int a = fb; a < lb; a++) {
        ;
        int lst = bck[a].back();
        bck[a].pop_back();
        int aa = a + 1;
        cnt[a][lst]--;
        cnt[aa][lst]++;
        bck[aa].push_front(lst);
      }
    } else {
      scanf("%d", &k);
      k = (k + last_ans - 1) % n + 1;
      int ans = 0;
      int fb = i / szb, lb = j / szb;
      for (int a = fb + 1; a < lb; a++) ans += cnt[a][k];
      int fs = i % szb, ls = j % szb;
      if (fb == lb) {
        for (int a = fs; a <= ls; a++) ans += (bck[fb][a] == k);
      } else {
        assert(fs < bck[fb].size());
        for (int a = fs; a < bck[fb].size(); a++) {
          ans += (bck[fb][a] == k);
        }
        assert(ls < bck[lb].size());
        for (int a = 0; a <= ls; a++) {
          ans += (bck[lb][a] == k);
        }
      }
      printf("%d\n", ans);
      last_ans = ans;
    }
  }
}