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 maxn = 1000005;
int a[maxn], b[maxn], ps[maxn], pos[maxn], ld, rd, x, n;
inline int f(const int z) { return (z - 1 + x) % n + 1; }
int *sq[maxn << 2];
void build(int l, int r, int rt) {
  sq[rt] = new int[r - l + 1];
  if (l == r) {
    sq[rt][0] = pos[l];
    return;
  }
  int i, m = (l + r) >> 1, p, q;
  build(l, m, rt << 1), build(m + 1, r, rt << 1 | 1);
  for (p = q = i = 0; p < m - l + 1 && q < r - m; i++) {
    if (sq[rt << 1][p] < sq[rt << 1 | 1][q])
      sq[rt][i] = sq[rt << 1][p++];
    else
      sq[rt][i] = sq[rt << 1 | 1][q++];
  }
  if (p == m - l + 1)
    for (; i < r - l + 1; i++) sq[rt][i] = sq[rt << 1 | 1][q++];
  else
    for (; i < r - l + 1; i++) sq[rt][i] = sq[rt << 1][p++];
}
int query(int L, int R, int l, int r, int rt) {
  if (L <= l && r <= R)
    return (upper_bound(sq[rt], sq[rt] + r - l + 1, rd) - sq[rt]) -
           (lower_bound(sq[rt], sq[rt] + r - l + 1, ld) - sq[rt]);
  int m = (l + r) >> 1, ret = 0;
  if (L <= m) ret += query(L, R, l, m, rt << 1);
  if (m < R) ret += query(L, R, m + 1, r, rt << 1 | 1);
  return ret;
}
int main() {
  int i, l, r, q, m;
  scanf("%d", &n);
  for (i = 0; i < n; i++) scanf("%d", &a[i]);
  for (i = 0; i < n; i++) scanf("%d", &b[i]), ps[b[i]] = i;
  for (i = 0; i < n; i++) pos[i] = ps[a[i]];
  build(0, n - 1, 1);
  scanf("%d", &m);
  for (x = 0; m--;) {
    int a, b, c, d;
    scanf("%d%d%d%d", &a, &b, &c, &d);
    l = f(a) - 1, r = f(b) - 1, ld = f(c) - 1, rd = f(d) - 1;
    if (l > r) swap(l, r);
    if (ld > rd) swap(ld, rd);
    printf("%d\n", query(l, r, 0, n - 1, 1));
    x++;
  }
  return 0;
}
