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 MN = 1000 * 1000 + 100;
const int lg = 30;
int r[MN];
int a[MN];
int seg[lg][MN];
int n;
int x = -1;
void build(int d = 0, int s = 0, int e = n) {
  if (e - s < 2) {
    seg[d][s] = a[s];
    return;
  }
  int mid = (s + e) / 2;
  build(d + 1, s, mid);
  build(d + 1, mid, e);
  merge(seg[d + 1] + s, seg[d + 1] + mid, seg[d + 1] + mid, seg[d + 1] + e,
        seg[d] + s);
}
int query(int l, int r, int k, int d = 0, int s = 0, int e = n) {
  if (s >= l && e <= r) {
    return upper_bound(seg[d] + s, seg[d] + e, k) - (seg[d] + s);
  }
  if (s >= r || l >= e) return 0;
  int mid = (s + e) / 2;
  return query(l, r, k, d + 1, s, mid) + query(l, r, k, d + 1, mid, e);
}
int f(int z) { return (z + x + n) % n; }
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    a[i]--;
  }
  for (int i = 0; i < n; i++) {
    int t;
    cin >> t;
    t--;
    r[t] = i;
  }
  for (int i = 0; i < n; i++) {
    a[i] = r[a[i]];
  }
  build();
  int m;
  cin >> m;
  for (int i = 0; i < m; i++) {
    int a, b, c, d;
    cin >> a >> b >> c >> d;
    a = f(a);
    b = f(b);
    if (a > b) swap(a, b);
    b++;
    c = f(c);
    d = f(d);
    if (c > d) swap(c, d);
    c--;
    cout << query(a, b, d) - query(a, b, c) << endl;
  }
  return 0;
}
