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;
static int A[1000010], B[1000010], tmp[1000010];
int N, l1, r1, l2, r2, ANS = 0, Q;
vector<vector<int> > ST(3000030);
inline void build(int x, int y, int n) {
  if (x == y) return void(ST[n].push_back(B[x]));
  int m = (x + y) >> 1;
  build(x, m, 2 * n);
  build(m + 1, y, 2 * n + 1);
  ST[n].resize(ST[2 * n].size() + ST[2 * n + 1].size());
  merge(ST[2 * n].begin(), ST[2 * n].end(), ST[2 * n + 1].begin(),
        ST[2 * n + 1].end(), back_inserter(ST[n]));
}
inline void query(int x, int y, int n) {
  if (y < l2 || r2 < x) return;
  if (l2 <= x && y <= r2) {
    int t = upper_bound(ST[n].begin(), ST[n].end(), r1) -
            lower_bound(ST[n].begin(), ST[n].end(), l1);
    ANS += t;
    return;
  }
  int m = (x + y) >> 1;
  query(x, m, 2 * n);
  query(m + 1, y, 2 * n + 1);
}
inline int f(int z) { return ((z - 1 + ANS) % N); }
int main() {
  scanf("%d", &N);
  for (int i = 1; i <= N; ++i) {
    scanf("%d", &A[i]);
    tmp[A[i]] = i;
  }
  for (int i = 1; i <= N; ++i) {
    scanf("%d", &B[i]);
    B[i] = tmp[B[i]];
  }
  build(1, N, 1);
  scanf("%d", &Q);
  while (Q--) {
    scanf("%d %d %d %d", &l1, &r1, &l2, &r2);
    l1 = min(f(l1), f(r1));
    r1 = max(f(l1), f(r1));
    l2 = min(f(l2), f(r2));
    r2 = max(f(l2), f(r2));
    ANS = 0;
    query(1, N, 1);
    printf("%d\n", ANS);
  }
  return 0;
}
