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 p[1100100], q[1100100];
int rp[1100100], rq[1100100];
int a[1100100];
struct segtree {
  int N;
  int dat[21][1048576];
  void init(int N_, int *a) {
    N = 1;
    int dep = 1;
    while (N < N_) {
      N *= 2;
      dep++;
    }
    for (int i = 0; i < N_; i++) dat[dep - 1][i] = a[i];
    for (int i = N_; i < N; i++) dat[dep - 1][i] = i;
    int ln = 1;
    for (int d = dep - 2; d >= 0; d--) {
      ln *= 2;
      for (int l = 0; l < N; l += ln) {
        int r = l + ln;
        int lid = l, rid = l + ln / 2;
        int id = l;
        while (lid < l + ln / 2 || rid < r) {
          if (lid == l + ln / 2) {
            dat[d][id] = dat[d + 1][rid];
            rid++;
            id++;
          } else if (rid == r) {
            dat[d][id] = dat[d + 1][lid];
            lid++;
            id++;
          } else {
            if (dat[d + 1][lid] < dat[d + 1][rid]) {
              dat[d][id] = dat[d + 1][lid];
              lid++;
              id++;
            } else {
              dat[d][id] = dat[d + 1][rid];
              rid++;
              id++;
            }
          }
        }
      }
    }
  }
  int query(int l, int r, int x, int d, int a, int b) {
    if (l <= a && b <= r)
      return upper_bound(&dat[d][a], &dat[d][b], x) - (&dat[d][a]);
    else if (r <= a || b <= l)
      return 0;
    else
      return query(l, r, x, d + 1, a, (a + b) / 2) +
             query(l, r, x, d + 1, (a + b) / 2, b);
  }
  int query(int l, int r, int x) {
    if (x < 0) return 0;
    return query(l, r, x, 0, 0, N);
  }
  void output() {
    for (int i = 0; i < 3; i++) {
      for (int j = 0; j < 4; j++) printf("%d ", dat[i][j]);
      printf("\n");
    }
  }
};
segtree seg;
int N;
int f(int x, int z) {
  int res = z - 1 + x;
  return (res % N) + 1;
}
int main() {
  scanf("%d", &N);
  for (int i = 0; i < N; i++) {
    scanf("%d", p + i);
    p[i]--;
    rp[p[i]] = i;
  }
  for (int i = 0; i < N; i++) {
    scanf("%d", q + i);
    q[i]--;
    rq[q[i]] = i;
  }
  for (int i = 0; i < N; i++) {
    int pos = rq[p[i]];
    a[i] = pos;
  }
  seg.init(N, a);
  int M;
  scanf("%d", &M);
  int x = 0;
  for (int i = 0; i < M; i++) {
    int a, b, c, d;
    scanf("%d%d%d%d", &a, &b, &c, &d);
    int l1, r1, l2, r2;
    l1 = min(f(x, a), f(x, b));
    r1 = max(f(x, a), f(x, b));
    l2 = min(f(x, c), f(x, d));
    r2 = max(f(x, c), f(x, d));
    l1--;
    l2--;
    r1--;
    r2--;
    int ll = seg.query(l2, r2 + 1, l1 - 1);
    int rr = seg.query(l2, r2 + 1, r1);
    printf("%d\n", rr - ll);
    x = rr - ll + 1;
  }
  return 0;
}
