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 long double eps = 1e-8;
int read() {
  int x = 0, f = 1;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (isdigit(c)) {
    x = (x << 3) + (x << 1) + (c ^ 48);
    c = getchar();
  }
  return x * f;
}
const int N = 1e6 + 10;
int a[N], b[N];
int val[N << 5], Ls[N << 5], Rs[N << 5], Rt[N];
int tot_seg;
int n, m;
int lst_ans;
inline int get_tr(int x) { return (x - 1 + lst_ans) % n + 1; }
int build(int l, int r) {
  int rt = ++tot_seg;
  val[rt] = 0;
  if (l == r) return rt;
  int mid = (l + r) >> 1;
  Ls[rt] = build(l, mid);
  Rs[rt] = build(mid + 1, r);
  return rt;
}
int modify(int pos, int l, int r, int x) {
  int rt = ++tot_seg;
  Ls[rt] = Ls[pos], Rs[rt] = Rs[pos], val[rt] = val[pos] + 1;
  if (l == r) return rt;
  int mid = (l + r) >> 1;
  if (x <= mid)
    Ls[rt] = modify(Ls[pos], l, mid, x);
  else
    Rs[rt] = modify(Rs[pos], mid + 1, r, x);
  return rt;
}
int query(int L, int R, int l, int r, int x) {
  if (L <= l && r <= R) return val[x];
  int mid = (l + r) >> 1, ret = 0;
  if (L <= mid) ret += query(L, R, l, mid, Ls[x]);
  if (mid < R) ret += query(L, R, mid + 1, r, Rs[x]);
  return ret;
}
signed main() {
  n = read();
  for (int i = (1), iE = (n); i <= iE; i++) b[read()] = i;
  for (int i = (1), iE = (n); i <= iE; i++) a[i] = b[read()];
  Rt[0] = build(1, n);
  for (int i = (1), iE = (n); i <= iE; i++)
    Rt[i] = modify(Rt[i - 1], 1, n, a[i]);
  m = read();
  while (m--) {
    int a = read(), b = read(), c = read(), d = read();
    int e = min(get_tr(a), get_tr(b)), f = max(get_tr(a), get_tr(b)),
        g = min(get_tr(c), get_tr(d)), h = max(get_tr(c), get_tr(d));
    printf("%d\n",
           lst_ans = query(e, f, 1, n, Rt[h]) - query(e, f, 1, n, Rt[g - 1]));
  }
  return 0;
}
