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 double eps = 1e-8;
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;
long long powmod(long long a, long long b) {
  long long res = 1;
  a %= mod;
  assert(b >= 0);
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
  }
  return res;
}
deque<int> q[510];
int p[333][100010];
int fl[1010], fr[1010];
int main() {
  int n, a;
  cin >> n;
  int m = sqrt(n) + 1;
  for (int i = 0; i < n; i++) {
    scanf("%d", &a);
    q[i / m].push_back(a);
    p[i / m][a]++;
  }
  int Q;
  cin >> Q;
  int la = 0, cnt, g;
  while (Q--) {
    int op, l, r;
    scanf("%d%d%d", &op, &l, &r);
    l = (l + la - 1) % n + 1;
    r = (r + la - 1) % n + 1;
    if (l > r) swap(l, r);
    l--;
    r--;
    int ll = l / m, rr = r / m;
    if (op == 1) {
      int cl = 0, cr = 0;
      cnt = ((int)(q[rr]).size()) - 1;
      while (cl + ll * m < l) {
        fl[cl++] = q[ll].front();
        q[ll].pop_front();
      }
      while (cnt + rr * m > r) {
        fr[cr++] = q[rr].back();
        q[rr].pop_back();
        cnt--;
      }
      g = q[rr].back();
      q[rr].pop_back();
      p[rr][g]--;
      p[ll][g]++;
      q[ll].push_front(g);
      for (int i = cl - 1; i >= 0; i--) q[ll].push_front(fl[i]);
      for (int i = cr - 1; i >= 0; i--) q[rr].push_back(fr[i]);
      for (int i = ll; i < rr; i++) {
        g = q[i].back();
        q[i].pop_back();
        p[i][g]--;
        p[i + 1][g]++;
        q[i + 1].push_front(g);
      }
    } else {
      int k;
      scanf("%d", &k);
      k = (k + la - 1) % n + 1;
      int ans = 0;
      for (int i = ll + 1; i < rr; i++) ans += p[i][k];
      cnt = 0;
      for (auto i : q[ll]) {
        if (cnt + ll * m >= l && cnt + ll * m <= r && i == k) ans++;
        cnt++;
      }
      if (ll != rr) {
        cnt = 0;
        for (auto i : q[rr]) {
          if (cnt + rr * m >= l && cnt + rr * m <= r && i == k) ans++;
          cnt++;
        }
      }
      printf("%d\n", ans);
      la = ans;
    }
  }
  return 0;
}