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 inds[1000000 + 1];
struct sgtr {
  struct vrtx {
    vector<int> sor = {};
    int sz = 0, ma = -((int)1e9 + 7), mi = ((int)1e9 + 7);
  };
  vrtx t[4 * 1000000];
  void build(int v, int l, int r, vector<int>& a) {
    if (l + 1 == r) {
      t[v].sor = {a[l]};
      t[v].ma = a[l];
      t[v].mi = a[l];
      t[v].sz = 1;
      return;
    }
    int mid = (l + r) >> 1;
    build(2 * v, l, mid, a);
    build(2 * v + 1, mid, r, a);
    t[v].ma = max(t[2 * v].ma, t[2 * v + 1].ma);
    t[v].mi = max(t[2 * v].mi, t[2 * v + 1].mi);
    int sz1 = t[2 * v].sz, sz2 = t[2 * v + 1].sz;
    t[v].sz = sz1 + sz2;
    int i1 = 0, i2 = 0;
    vector<int> fir = t[2 * v].sor, sec = t[2 * v + 1].sor;
    while (i1 < sz1 || i2 < sz2) {
      while (i1 == sz1 && i2 < sz2) {
        t[v].sor.push_back(sec[i2]);
        ++i2;
      }
      while (i2 == sz2 && i1 < sz1) {
        t[v].sor.push_back(fir[i1]);
        ++i1;
      }
      while (i2 < sz2 && i1 < sz1) {
        if (fir[i1] <= sec[i2]) {
          t[v].sor.push_back(fir[i1]);
          ++i1;
          continue;
        }
        t[v].sor.push_back(sec[i2]);
        ++i2;
      }
    }
  }
  int gmore_x(int v, int l, int r, int l_q, int r_q, int x) {
    if (l >= r_q || r <= l_q) return 0;
    if (l >= l_q && r <= r_q) {
      return (t[v].sz - (lower_bound(t[v].sor.begin(), t[v].sor.end(), x) -
                         t[v].sor.begin()));
    }
    int mid = (l + r) >> 1;
    return gmore_x(2 * v, l, mid, l_q, r_q, x) +
           gmore_x(2 * v + 1, mid, r, l_q, r_q, x);
  }
};
int x = 0;
int f(int z, int n) { return ((z - 1 + x) % n + 1); }
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int n;
  cin >> n;
  vector<int> A(n), B(n), a(n);
  for (int i = 0; i < n; ++i) {
    cin >> A[i];
    inds[A[i]] = i;
  }
  for (int i = 0; i < n; ++i) {
    cin >> B[i];
    a[inds[B[i]]] = i;
  }
  sgtr now;
  now.build(1, 0, n, a);
  int m;
  cin >> m;
  for (int i = 0; i < m; ++i) {
    int a, b, c, d;
    cin >> a >> b >> c >> d;
    int l1 = min(f(a, n), f(b, n)), r1 = max(f(a, n), f(b, n)),
        l2 = min(f(c, n), f(d, n)), r2 = max(f(c, n), f(d, n));
    --l1;
    --l2;
    int fir = now.gmore_x(1, 0, n, l1, r1, l2);
    int sec = now.gmore_x(1, 0, n, l1, r1, r2 + 1);
    x = fir - sec;
    cout << x << '\n';
    ++x;
  }
}
