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 long long inf = 4e18;
const int N = 700000;
const int s = 500;
int n, m, a[N], b[460][120000], res;
deque<int> q[N], q1;
void change(int* val) { *val = (*val + res - 1) % n; }
int main() {
  scanf("%d", &n);
  for (int i = 0; i <= n - 1; i++) scanf("%d", &a[i]);
  for (int i = 0; i <= n - 1; i++) b[i / s][a[i]]++;
  for (int i = 0; i <= n - 1; i++) q[i / s].push_back(a[i]);
  scanf("%d", &m);
  res = 0;
  while (m--) {
    int type, l, r;
    scanf("%d", &type);
    if (type == 1) {
      scanf("%d%d", &l, &r);
      change(&l);
      change(&r);
      if (l > r) swap(l, r);
      if (l == r) continue;
      int i1 = l / s, i2 = r / s;
      for (int i = i1; i <= i2 - 1; i++) {
        b[i + 1][q[i].back()]++;
        b[i][q[i].back()]--;
        q[i + 1].push_front(q[i].back());
        q[i].pop_back();
      }
      int f = 0;
      if (i1 != i2) f = 1;
      int val;
      int kol = 0;
      for (deque<int>::iterator it = q[i2].begin(); it != q[i2].end(); it++) {
        if (kol == r % s + f) {
          val = (*it);
          break;
        }
        kol++;
      }
      q[i2].erase(q[i2].begin() + r % s + f);
      q[i1].insert(q[i1].begin() + l % s, val);
      b[i2][val]--;
      b[i1][val]++;
    } else {
      int k;
      scanf("%d%d%d", &l, &r, &k);
      change(&l);
      change(&r);
      change(&k);
      k++;
      if (l > r) swap(l, r);
      int i1 = l / s, i2 = r / s;
      res = 0;
      for (int i = i1 + 1; i <= i2 - 1; i++) res += b[i][k];
      int kol = 0;
      for (deque<int>::iterator it = q[i1].begin(); it != q[i1].end(); it++) {
        if (i1 != i2) {
          if (kol >= l % s && (*it) == k) res++;
        } else {
          if (kol >= l % s && kol <= r % s && (*it) == k) res++;
        }
        kol++;
      }
      if (i1 != i2) {
        kol = 0;
        for (deque<int>::iterator it = q[i2].begin(); it != q[i2].end(); it++) {
          if (kol <= r % s && (*it) == k) res++;
          kol++;
        }
      }
      printf("%d\n", res);
    }
  }
  return 0;
}