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;
inline int read() {
  int f = 1, ans = 0;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    ans = ans * 10 + c - '0';
    c = getchar();
  }
  return ans * f;
}
const int MAXN = 1000001;
int N, X[MAXN], Y[MAXN], rt[MAXN];
vector<int> vec[MAXN];
struct Segment {
  int tot, ls[MAXN << 4], rs[MAXN << 4], Ans[MAXN << 4];
  void build(int &rt, int l, int r) {
    rt = ++tot;
    if (l == r) return;
    int mid = l + r >> 1;
    build(ls[rt], l, mid), build(rs[rt], mid + 1, r);
    return;
  }
  void Modify(int &p, int q, int l, int r, int ps) {
    p = ++tot;
    ls[p] = ls[q], rs[p] = rs[q], Ans[p] = Ans[q] + 1;
    if (l == r) return;
    int mid = l + r >> 1;
    if (ps <= mid) Modify(ls[p], ls[q], l, mid, ps);
    if (mid < ps) Modify(rs[p], rs[q], mid + 1, r, ps);
    return;
  }
  int Query(int k, int l, int r, int x, int y) {
    if (x <= l && r <= y) return Ans[k];
    int mid = l + r >> 1, res = 0;
    if (x <= mid) res += Query(ls[k], l, mid, x, y);
    if (mid < y) res += Query(rs[k], mid + 1, r, x, y);
    return res;
  }
} T;
int lasans;
int f(int x) { return (x - 1 + lasans) % N + 1; }
int main() {
  N = read();
  for (int i = 1; i <= N; i++) X[read()] = i;
  for (int i = 1; i <= N; i++) Y[read()] = i;
  for (int i = 1; i <= N; i++) vec[X[i]].push_back(Y[i]);
  T.build(rt[0], 1, N);
  for (int i = 1; i <= N; i++) {
    rt[i] = rt[i - 1];
    for (int j = 0; j < vec[i].size(); j++)
      T.Modify(rt[i], rt[i], 1, N, vec[i][j]);
  }
  int Q = read();
  for (int i = 1; i <= Q; i++) {
    int x1 = f(read()), x2 = f(read()), y1 = f(read()), y2 = f(read());
    if (x1 > x2) swap(x1, x2);
    if (y1 > y2) swap(y1, y2);
    printf("%d\n", lasans = T.Query(rt[x2], 1, N, y1, y2) -
                            T.Query(rt[x1 - 1], 1, N, y1, y2));
    lasans++;
  }
  return 0;
}
