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;
int cnt[320][100010];
int val[320][320];
int blen[320];
int org[320];
int main() {
  int n;
  cin >> n;
  int blend = int(sqrt(n));
  int nb = 0;
  int sisa = n;
  while (sisa) {
    blen[nb] = min(sisa, blend);
    sisa -= blen[nb];
    nb += 1;
  }
  int bucket = 0;
  int pos = 0;
  for (int i = 0; i < n; ++i) {
    int a;
    cin >> a;
    val[bucket][pos] = a;
    cnt[bucket][a] += 1;
    pos += 1;
    if (pos == blen[bucket]) {
      bucket += 1;
      pos = 0;
    }
  }
  int q;
  cin >> q;
  int lastans = 0;
  for (int(_) = 0; (_) < (q); ++(_)) {
    int x;
    int l, r;
    scanf("%d%d%d", &x, &l, &r);
    l = ((l + lastans - 1) % n);
    r = ((r + lastans - 1) % n);
    if (l > r) swap(l, r);
    if (x == 1) {
      int firstbucket = l / blend;
      int lastbucket = r / blend;
      if (firstbucket == lastbucket) {
        l %= blend;
        r %= blend;
        int modu = blen[firstbucket];
        int off = org[firstbucket];
        int lasty = val[firstbucket][(off + r) % modu];
        for (int i = r; i > l; --i) {
          val[firstbucket][(off + i) % modu] =
              val[firstbucket][(off + i + modu - 1) % modu];
        }
        val[firstbucket][(off + l) % modu] = lasty;
      } else {
        l %= blend;
        r %= blend;
        int carry = val[lastbucket][(org[lastbucket] + r) % blen[lastbucket]];
        for (int i = firstbucket; i <= lastbucket; ++i) {
          int modu = blen[i];
          int off = org[i];
          if (i == firstbucket || i == lastbucket) {
            int lo = l;
            int hi = blen[i] - 1;
            if (i == lastbucket) {
              lo = 0;
              hi = r;
            }
            int nc = val[i][(off + hi) % modu];
            for (int j = hi; j > lo; --j) {
              val[i][(off + j) % modu] = val[i][(off + j + modu - 1) % modu];
            }
            val[i][(off + lo) % modu] = carry;
            cnt[i][nc] -= 1;
            cnt[i][carry] += 1;
            carry = nc;
          } else {
            org[i] = (off - 1 + modu) % modu;
            int nc = val[i][org[i]];
            val[i][org[i]] = carry;
            cnt[i][nc] -= 1;
            cnt[i][carry] += 1;
            carry = nc;
          }
        }
      }
    } else {
      int k;
      scanf("%d", &k);
      k = 1 + ((k + lastans - 1) % n);
      int jml = 0;
      int res = 0;
      for (int(i) = 0; (i) < (nb); ++(i)) {
        int kan = jml + blen[i] - 1;
        if (l <= jml && r >= kan) {
          res += cnt[i][k];
        } else if (l > kan || r < jml) {
        } else {
          for (int(j) = 0; (j) < (blen[i]); ++(j))
            if (jml + j >= l && jml + j <= r &&
                val[i][(org[i] + j) % blen[i]] == k)
              ++res;
        }
        jml += blen[i];
      }
      cout << res << endl;
      lastans = res;
    }
  }
  return 0;
}