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 N = 1e6 + 2;
int n, adr[N], q[N], x;
vector<int> t[4 * N];
void build(int n, int b, int e) {
  if (b == e) {
    t[n].push_back(adr[q[b]]);
    return;
  }
  int l = n << 1, r = l | 1, m = b + e >> 1;
  build(l, b, m);
  build(r, m + 1, e);
  int p1 = 0, p2 = 0;
  while (p1 < t[l].size() && p2 < t[r].size())
    if (t[l][p1] < t[r][p2])
      t[n].push_back(t[l][p1++]);
    else
      t[n].push_back(t[r][p2++]);
  while (p1 < t[l].size()) t[n].push_back(t[l][p1++]);
  while (p2 < t[r].size()) t[n].push_back(t[r][p2++]);
}
int query(int n, int b, int e, int u, int v, int x, int y) {
  if (e < u || v < b) return 0;
  if (u <= b && e <= v)
    return upper_bound(t[n].begin(), t[n].end(), y) -
           lower_bound(t[n].begin(), t[n].end(), x);
  int l = n << 1, r = l | 1, m = b + e >> 1;
  return query(l, b, m, u, v, x, y) + query(r, m + 1, e, u, v, x, y);
}
inline int f(int z) { return (z + x) % n; }
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  for (int i = 0, t; i < n; i++) cin >> t, adr[--t] = i;
  for (int i = 0; i < n; i++) cin >> q[i], q[i]--;
  build(1, 0, n - 1);
  int m;
  cin >> m;
  for (int i = 0; i < m; i++, x++) {
    int b, e, l, r;
    cin >> b >> e >> l >> r;
    b = f(--b), e = f(--e), l = f(--l), r = f(--r);
    cout << (x = query(1, 0, n - 1, min(l, r), max(l, r), min(b, e), max(b, e)))
         << '\n';
  }
  return 0;
}