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 main() {
  int n;
  int maxn = 1000001;
  int p[maxn];
  int q[maxn];
  cin >> n;
  cin.get();
  for (int i = 1; i <= n; ++i) cin >> p[i];
  cin.get();
  for (int i = 1; i <= n; ++i) cin >> q[i];
  cin.get();
  map<int, pair<int, int> > r;
  for (int i = 1; i <= n; ++i) {
    r[i] = make_pair(0, 0);
  }
  for (int i = 1; i <= n; ++i) {
    r[p[i]].first = i;
    r[q[i]].second = i;
  }
  int nq;
  cin >> nq;
  cin.get();
  int a, b, c, d;
  long long x = 0;
  int l1, r1, l2, r2;
  long long prev = 0;
  for (int i = 0; i < nq; ++i) {
    cin >> a >> b >> c >> d;
    cin.get();
    if (i == 0)
      x = 0;
    else
      x = prev + 1;
    l1 = min((a - 1 + x) % n + 1, (b - 1 + x) % n + 1);
    r1 = max((a - 1 + x) % n + 1, (b - 1 + x) % n + 1);
    l2 = min((c - 1 + x) % n + 1, (d - 1 + x) % n + 1);
    r2 = max((c - 1 + x) % n + 1, (d - 1 + x) % n + 1);
    int cnt = 0;
    for (int i = l1; i <= r1; ++i) {
      if (r[i].second >= l2 && r[i].second <= r2) ++cnt;
    }
    prev = cnt;
    cout << cnt << endl;
  }
  return 0;
}
