Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <typename TH>
void _dbg(const char* sdbg, TH h) {
  cerr << sdbg << " = " << h << "\n";
}
template <typename TH, typename... TA>
void _dbg(const char* sdbg, TH h, TA... t) {
  while (*sdbg != ',') cerr << *sdbg++;
  cerr << " = " << h << ",";
  _dbg(sdbg + 1, t...);
}
const int N = 1e6 + 6;
int n, a[N], b[N], q;
vector<int> st[4 * N];
void Build(int id, int l, int r) {
  if (l == r) {
    st[id].push_back(a[l]);
    return;
  }
  Build(id * 2, l, (l + r) / 2);
  Build(id * 2 + 1, (l + r) / 2 + 1, r);
  merge(begin(st[id * 2]), end(st[id * 2]), begin(st[id * 2 + 1]),
        end(st[id * 2 + 1]), back_inserter(st[id]));
}
int Count(int id, int l, int r, int i, int j, int x, int y) {
  if (r < i || l > j) {
    return 0;
  }
  if (i <= l && r <= j) {
    int pos1 = upper_bound(begin(st[id]), end(st[id]), y) - begin(st[id]) - 1;
    int pos2 = lower_bound(begin(st[id]), end(st[id]), x) - begin(st[id]);
    if (pos1 >= pos2) {
      return pos1 - pos2 + 1;
    }
    return 0;
  }
  return Count(id * 2, l, (l + r) / 2, i, j, x, y) +
         Count(id * 2 + 1, (l + r) / 2 + 1, r, i, j, x, y);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
  }
  for (int i = 1; i <= n; ++i) {
    int x;
    cin >> x;
    b[x] = i;
  }
  for (int i = 1; i <= n; ++i) {
    a[i] = b[a[i]];
  }
  Build(1, 1, n);
  int x = 0;
  auto F = [&x](int z) { return ((z - 1 + x) % n) + 1; };
  cin >> q;
  while (q--) {
    int A, B, C, D;
    cin >> A >> B >> C >> D;
    int l1 = min(F(A), F(B)), r1 = max(F(A), F(B)), l2 = min(F(C), F(D)),
        r2 = max(F(C), F(D));
    x = Count(1, 1, n, l1, r1, l2, r2);
    cout << x << '\n';
    x++;
  }
  return 0;
}