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 = 1000100;
int n;
int a[N], b[N];
int pos[N];
int tot = 0;
int rt[N], ls[N * 20], rs[N * 20], sum[N * 20];
void build(int l, int r, int& rt) {
  rt = tot++;
  sum[rt] = 0;
  if (l == r) return;
  int m = (l + r) >> 1;
  build(l, m, ls[rt]);
  build(m + 1, r, rs[rt]);
}
void update(int pos, int l, int r, int last, int& rt) {
  rt = tot++;
  ls[rt] = ls[last];
  rs[rt] = rs[last];
  sum[rt] = sum[last] + 1;
  if (l == r) return;
  int m = (l + r) >> 1;
  if (pos <= m)
    update(pos, l, m, ls[last], ls[rt]);
  else
    update(pos, m + 1, r, rs[last], rs[rt]);
}
int query(int L, int R, int l, int r, int last, int rt) {
  if (L <= l && r <= R) return sum[rt] - sum[last];
  int m = (l + r) >> 1;
  if (R <= m)
    return query(L, R, l, m, ls[last], ls[rt]);
  else if (L > m)
    return query(L, R, m + 1, r, rs[last], rs[rt]);
  else
    return query(L, m, l, m, ls[last], ls[rt]) +
           query(m + 1, R, m + 1, R, rs[last], rs[rt]);
}
int f(int z, int x) { return ((z - 1 + x) % n) + 1; }
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= n; i++) {
    cin >> b[i];
    pos[b[i]] = i;
  }
  for (int i = 1; i <= n; i++) a[i] = pos[a[i]];
  build(1, n, rt[0]);
  for (int i = 1; i <= n; i++) {
    update(a[i], 1, n, rt[i - 1], rt[i]);
  }
  int m, l1, r1, l2, r2, ans = 0;
  cin >> m;
  while (m--) {
    int a, b, c, d;
    cin >> a >> b >> c >> d;
    l1 = min(f(a, ans), f(b, ans));
    r1 = max(f(a, ans), f(b, ans));
    l2 = min(f(c, ans), f(d, ans));
    r2 = max(f(c, ans), f(d, ans));
    ans = query(l2, r2, 1, n, rt[l1 - 1], rt[r1]);
    cout << ans << endl;
  }
  return 0;
}
