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 a[1000010], b[1000010];
int aa[1000010], bb[1000010];
int root[1000010], nodesize = 0;
int sum[1000010 * 20];
int lson[1000010 * 20], rson[1000010 * 20];
int build(int l, int r) {
  int x = ++nodesize;
  sum[x] = lson[x] = rson[x] = 0;
  if (l == r) return x;
  int mid = l + (r - l) / 2;
  lson[x] = build(l, mid);
  rson[x] = build(mid + 1, r);
  return x;
}
int k;
int update(int l, int r, int preo) {
  int x = ++nodesize;
  lson[x] = lson[preo];
  rson[x] = rson[preo];
  sum[x] = sum[preo] + 1;
  if (l == r) return x;
  int mid = l + (r - l) / 2;
  if (k <= mid)
    lson[x] = update(l, mid, lson[preo]);
  else
    rson[x] = update(mid + 1, r, rson[preo]);
  return x;
}
int ql, qr;
int query(int l, int r, int u, int v) {
  if (ql <= l && r <= qr) return sum[v] - sum[u];
  int mid = l + (r - l) / 2;
  int sum = 0;
  if (ql <= mid)
    sum += query(l, mid, lson[u], lson[v]);
  else
    sum += query(mid + 1, r, rson[u], rson[v]);
  return sum;
}
int n;
inline int F(int z, int x) { return (z - 1 + x) % n + 1; }
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    aa[i] = a[i];
  }
  for (int i = 1; i <= n; i++) {
    scanf("%d", &b[i]);
    bb[b[i]] = i;
  }
  for (int i = 1; i <= n; i++) a[i] = bb[aa[i]];
  root[0] = build(1, n);
  for (int i = 1; i <= n; i++) {
    k = a[i];
    root[i] = update(1, n, root[i - 1]);
  }
  int m, x = 0;
  scanf("%d", &m);
  for (int i = 1; i <= m; i++) {
    int a, b, c, d;
    scanf("%d%d%d%d", &a, &b, &c, &d);
    int l, r;
    l = F(a, x);
    r = F(b, x);
    ql = F(c, x);
    qr = F(d, x);
    if (l > r) swap(l, r);
    if (ql > qr) swap(ql, qr);
    int ans = query(1, n, root[l - 1], root[r]);
    printf("%d\n", ans);
    x = ans + 1;
  }
  return 0;
}
