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;
const int MAXN = 1e6 + 5;
int ch[MAXN], arr[MAXN], x, a, b, c, d, l1, l2, r1, r2, n, m;
vector<int> tree[4 * MAXN];
void build(int v, int b, int e) {
  if (b == e)
    tree[v].push_back(arr[b]);
  else {
    int m = (b + e) / 2;
    build(v + v, b, m);
    build(v + v + 1, m + 1, e);
    vector<int> &l = tree[v + v], &r = tree[v + v + 1];
    merge(l.begin(), l.end(), r.begin(), r.end(), back_inserter(tree[v]));
  }
}
int query(int v, int b, int e, int l, int r, int f, int s) {
  if (e < l || b > r) return 0;
  if (l <= b && e <= r) {
    return upper_bound(tree[v].begin(), tree[v].end(), s) -
           lower_bound(tree[v].begin(), tree[v].end(), f);
  }
  int m = (b + e) / 2;
  return query(v + v, b, m, l, r, f, s) +
         query(v + v + 1, m + 1, e, l, r, f, s);
}
int f(int z) { return ((z - 1 + x) % n) + 1; }
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> a;
    ch[a] = i + 1;
  }
  for (int i = 1; i <= n; ++i) {
    cin >> a;
    arr[i] = ch[a];
  }
  build(1, 1, n);
  cin >> m;
  for (int i = 0; i < m; ++i) {
    cin >> a >> b >> c >> d;
    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 = query(1, 1, n, l2, r2, l1, r1) + 1;
    cout << x - 1 << '\n';
  }
  return 0;
}