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;
int son[1110100 * 21][2], S[1110100 * 21], root[1110100], a[1110100],
    b[1110100], tmp[1110100];
int n, m, cc = 0, ans;
inline int read() {
  int t = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') t = t * 10 + ch - '0', ch = getchar();
  return t * f;
}
inline void Add(int la, int &p, int l, int r, int x) {
  p = ++cc;
  son[p][0] = son[la][0], son[p][1] = son[la][1], S[p] = S[la] + 1;
  if (l == r) return;
  int mid = (l + r) >> 1;
  if (x <= mid)
    Add(son[la][0], son[p][0], l, mid, x);
  else
    Add(son[la][1], son[p][1], mid + 1, r, x);
}
inline void query(int la, int p, int l, int r, int s, int t) {
  if (l == s && r == t) {
    ans += S[p] - S[la];
    return;
  }
  int mid = (l + r) >> 1;
  if (t <= mid)
    query(son[la][0], son[p][0], l, mid, s, t);
  else if (s > mid)
    query(son[la][1], son[p][1], mid + 1, r, s, t);
  else
    query(son[la][0], son[p][0], l, mid, s, mid),
        query(son[la][1], son[p][1], mid + 1, r, mid + 1, t);
}
int main() {
  n = read();
  for (int i = 1; i <= n; i++) a[i] = read();
  for (int i = 1; i <= n; i++) b[i] = read();
  for (int i = 1; i <= n; i++) tmp[b[i]] = i;
  for (int i = 1; i <= n; i++) Add(root[i - 1], root[i], 1, n, tmp[a[i]]);
  m = read();
  int l1, l2, r1, r2, t1, t2;
  ans = 0;
  for (int T = 1; T <= m; T++) {
    l1 = read(), r1 = read(), t1 = ((l1 - 1 + ans) % n + 1),
    t2 = ((r1 - 1 + ans) % n + 1), l1 = min(t1, t2), r1 = max(t1, t2);
    l2 = read(), r2 = read(), t1 = ((l2 - 1 + ans) % n + 1),
    t2 = ((r2 - 1 + ans) % n + 1), l2 = min(t1, t2), r2 = max(t1, t2);
    ans = 0;
    query(root[l1 - 1], root[r1], 1, n, l2, r2);
    printf("%d\n", ans);
    ans++;
  }
  return 0;
}
