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;
const int N = 1000005, M = 6005, sz = 200;
int n, m, c, p[N], q[N], a[N], b[N], z[M][M], s[M][M], x;
int f(int z) { return (z - 1 + x) % n + 1; }
int sol(int h, int v) {
  int ans = s[h / sz][v / sz];
  for (int i = h / sz * sz; i < h; i++)
    if (b[i] < v) ans++;
  for (int i = v / sz * sz; i < v; i++)
    if (a[i] < h / sz * sz) ans++;
  return ans;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &p[i]);
  for (int i = 0; i < n; i++) scanf("%d", &q[i]);
  for (int i = 0; i < n; i++) a[p[i]] = i;
  for (int i = 0; i < n; i++) b[a[q[i]]] = i;
  for (int i = 0; i < n; i++) a[b[i]] = i;
  for (int i = 0; i < n; i++) s[i / sz][a[i] / sz]++;
  for (int i = 0; i < M - 1; i++)
    for (int j = 0; j < M - 1; j++)
      s[i + 1][j + 1] += s[i + 1][j] + s[i][j + 1] - s[i][j];
  scanf("%d", &m);
  while (m--) {
    int l1, r1, l2, r2;
    scanf("%d%d%d%d", &l1, &r1, &l2, &r2);
    l1 = f(l1), r1 = f(r1), l2 = f(l2), r2 = f(r2);
    l1--, r1--, l2--, r2--;
    if (l1 > r1) swap(l1, r1);
    if (l2 > r2) swap(l2, r2);
    x = sol(r1 + 1, r2 + 1) - sol(l1, r2 + 1) - sol(r1 + 1, l2) + sol(l1, l2);
    printf("%d\n", x);
    x++;
  }
  return 0;
}
