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 int maxn = 1000000 + 10;
int n, a[maxn], b[maxn], m, pos[maxn], g;
vector<int> sum[maxn * 4], flag[maxn * 4];
void add(int root, int lo, int hi, int P, int c) {
  if (P > hi || lo > P) return;
  if (P <= lo && P >= hi) {
    flag[root].push_back(c);
    sum[root].push_back(c);
    return;
  }
  int mid = (lo + hi) / 2;
  add(2 * root, lo, mid, P, c);
  add(2 * root + 1, mid + 1, hi, P, c);
  sum[root].push_back(c);
}
int get(int root, int lo, int hi, int L, int R, int i, int j) {
  if (L > hi || lo > R) return 0;
  if (L <= lo && R >= hi) {
    int l =
        lower_bound(sum[root].begin(), sum[root].end(), i) - sum[root].begin();
    int r =
        upper_bound(sum[root].begin(), sum[root].end(), j) - sum[root].begin();
    return r - l;
  }
  int mid = (lo + hi) / 2;
  return get(2 * root, lo, mid, L, R, i, j) +
         get(2 * root + 1, mid + 1, hi, L, R, i, j);
}
void push(int root, int lo, int hi) {
  if (lo == hi) return;
  for (int i = 0; i < flag[root].size(); i++) {
    sum[2 * root].push_back(flag[root][i]);
    sum[2 * root + 1].push_back(flag[root][i]);
  }
  int mid = (lo + hi) / 2;
  push(2 * root, lo, mid);
  push(2 * root + 1, mid + 1, hi);
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    pos[a[i]] = i;
  }
  for (int i = 1; i <= n; i++) {
    cin >> b[i];
    add(1, 1, n, pos[b[i]], i);
  }
  push(1, 1, n);
  for (int i = 1; i < 4 * maxn; i++) sort(sum[i].begin(), sum[i].end());
  cin >> m;
  for (int i = 0; i < m; i++) {
    int x1, y1, x2, y2, a1, b1, a2, b2;
    cin >> x1 >> y1 >> x2 >> y2;
    a1 = 1 + min((x1 + i - 1) % n, (y1 + i - 1) % n);
    b1 = 1 + max((x1 + i - 1) % n, (y1 + i - 1) % n);
    a2 = 1 + min((x2 + i - 1) % n, (y2 + i - 1) % n);
    b2 = 1 + max((x2 + i - 1) % n, (y2 + i - 1) % n);
    cout << get(1, 1, n, a1, b1, a2, b2) << endl;
  }
  return 0;
}
