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>
bool verifica(long int x, long int p[1000000], long int n) {
  for (long int i = 0; i < n; i++) {
    if (p[i] == x) {
      return true;
    }
  }
  return false;
}
int main() {
  long int i, n, m, a, b, c, d, p[100000], q[100000];
  scanf("%ld", &n);
  for (i = 0; i < n; i++) {
    scanf("%ld", &p[i]);
  }
  for (i = 0; i < n; i++) {
    scanf("%ld", &q[i]);
  }
  scanf("%ld", &m);
  for (long int j = 0; j < m; j++) {
    scanf("%ld %ld %ld %ld", &a, &b, &c, &d);
    long int nr = 0, nr1, nr2;
    if (b - a < 0 || d - c < 0) {
      printf("1\n");
    } else {
      nr1 = b - a + 1;
      nr2 = d - c + 1;
      if (nr1 > n / 2) {
        nr = n;
        for (i = 0; i < a; i++) {
          if (verifica(p[i], q, n)) {
            nr--;
          }
        }
        for (i = b; i < n; i++) {
          if (verifica(p[i], q, n)) {
            nr--;
          }
        }
      } else {
        for (i = a - 1; i < b; i++) {
          if (verifica(p[i], q, n)) {
            nr++;
          }
        }
        nr--;
      }
      printf("%ld\n", nr);
    }
  }
  return 0;
}
