You are given two permutations p and q, consisting of n elements, and m queries of the form: l1, r1, l2, r2 (l1 ≤ r1; l2 ≤ r2). The response for the query is the number of such integers from 1 to n, that their position in the first permutation is in segment [l1, r1] (borders included), and position in the second permutation is in segment [l2, r2] (borders included too).

A permutation of n elements is the sequence of n distinct integers, each not less than 1 and not greater than n.

Position of number v (1 ≤ v ≤ n) in permutation g1, g2, ..., gn is such number i, that gi = v.

Input

The first line contains one integer n (1 ≤ n ≤ 106), the number of elements in both permutations. The following line contains n integers, separated with spaces: p1, p2, ..., pn (1 ≤ pi ≤ n). These are elements of the first permutation. The next line contains the second permutation q1, q2, ..., qn in same format.

The following line contains an integer m (1 ≤ m ≤ 2·105), that is the number of queries.

The following m lines contain descriptions of queries one in a line. The description of the i-th query consists of four integers: a, b, c, d (1 ≤ a, b, c, d ≤ n). Query parameters l1, r1, l2, r2 are obtained from the numbers a, b, c, d using the following algorithm: 

  1. Introduce variable x. If it is the first query, then the variable equals 0, else it equals the response for the previous query plus one. 
  2. Introduce function f(z) = ((z - 1 + x) mod n) + 1. 
  3. Suppose l1 = min(f(a), f(b)), r1 = max(f(a), f(b)), l2 = min(f(c), f(d)), r2 = max(f(c), f(d)). 

Output

Print a response for each query in a separate line.

Examples

Input

3
3 1 2
3 2 1
1
1 2 3 3


Output

1


Input

4
4 3 2 1
2 3 4 1
3
1 2 3 4
1 3 2 1
1 4 2 3


Output

1
1
2
1
1
1
1
1 1 1 1

1

#include <bits/stdc++.h>
using namespace std;
int N, M, P[1111111], Q[1111111], z[4], x;
int l1, r1, l2, r2;
vector<int> Num[1111111];
void updata(int n, int x) {
  while (n <= N) {
    Num[n].push_back(x);
    n += n & (-n);
  }
}
int cnt(int n, int x) {
  int ret = 0;
  while (n >= 1) {
    ret += upper_bound(Num[n].begin(), Num[n].end(), x) - Num[n].begin();
    n -= n & (-n);
  }
  return ret;
}
int main() {
  scanf("%d", &N);
  for (int n = 1; n <= N; ++n) {
    scanf("%d", &x);
    P[x] = n;
  }
  for (int n = 1; n <= N; ++n) {
    scanf("%d", &x);
    Q[x] = n;
  }
  for (int n = 1; n <= N; ++n) updata(P[n], Q[n]);
  for (int n = 1; n <= N; ++n) sort(Num[n].begin(), Num[n].end());
  scanf("%d", &M);
  x = 0;
  while (M--) {
    for (int d = 0; d < 4; ++d) {
      scanf("%d", z + d);
      z[d] = (z[d] - 1 + x) % N + 1;
    }
    l1 = min(z[0], z[1]);
    r1 = max(z[0], z[1]);
    l2 = min(z[2], z[3]);
    r2 = max(z[2], z[3]);
    x = cnt(r1, r2) - cnt(r1, l2 - 1) - cnt(l1 - 1, r2) + cnt(l1 - 1, l2 - 1);
    printf("%d\n", x++);
  }
  return 0;
}
