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>
int func(int param, int x, int n) { return ((param - 1 + x) % n) + 1; }
int max(int a, int b) {
  if (a > b) return a;
  return b;
}
int min(int a, int b) {
  if (a > b) return b;
  return a;
}
int count_equal_numbers(int* perm1, int* perm2, int l1, int r1, int l2,
                        int r2) {
  int d = 0;
  for (int i = l1; i <= r1; ++i) {
    for (int j = l2; j <= r2; ++j) {
      if (perm1[i] == perm2[j]) {
        d = d + 1;
      }
    }
  }
  return d;
}
int exec_query(int* perm1, int* perm2, int query[], int x, int n) {
  int l1 = min(func(query[0], x, n), func(query[1], x, n));
  int r1 = max(func(query[0], x, n), func(query[1], x, n));
  int l2 = min(func(query[2], x, n), func(query[3], x, n));
  int r2 = max(func(query[2], x, n), func(query[3], x, n));
  return count_equal_numbers(perm1, perm2, l1, r1, l2, r2);
}
int main() {
  int n, m;
  int *perm1, *perm2;
  scanf("%d", &n);
  perm1 = (int*)malloc(n * sizeof(int) + 1);
  perm2 = (int*)malloc(n * sizeof(int) + 1);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &perm1[i]);
  }
  for (int i = 1; i <= n; i++) {
    scanf("%d", &perm2[i]);
  }
  scanf("%d", &m);
  int queries[m][4];
  for (int i = 0; i < m; ++i) {
    for (int j = 0; j < 4; ++j) {
      scanf("%d", &queries[i][j]);
    }
  }
  int x = 0;
  for (int i = 0; i < m; i++) {
    int digs = exec_query(perm1, perm2, queries[i], x, n);
    x += digs;
    printf("%d\n", digs);
  }
  free(perm1);
  free(perm2);
}
