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 + 5;
int n, x, a[N], b[N], ans;
vector<int> t[4 * N];
inline int f(int z) { return ((z - 1 + ans) % n + n) % n + 1; }
void build(int v, int l, int r) {
  if (l == r) {
    t[v].push_back(a[l]);
    return;
  }
  int mid = (r + l) >> 1;
  build(v + v, l, mid);
  build(v + v + 1, mid + 1, r);
  merge(t[v + v].begin(), t[v + v].end(), t[v + v + 1].begin(),
        t[v + v + 1].end(), back_inserter(t[v]));
}
void get(int v, int l, int r, int tl, int tr, int x, int y) {
  if (l > r || l > tr || tl > r) {
    return;
  }
  if (tl <= l && r <= tr) {
    int val = int(upper_bound(t[v].begin(), t[v].end(), y) -
                  lower_bound(t[v].begin(), t[v].end(), x));
    ans += val;
    return;
  }
  int mid = (r + l) >> 1;
  get(v + v, l, mid, tl, tr, x, y);
  get(v + v + 1, mid + 1, r, tl, tr, x, y);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> x;
    b[x] = i;
  }
  for (int i = 1; i <= n; i++) {
    cin >> x;
    a[i] = b[x];
  }
  build(1, 1, n);
  int m;
  cin >> m;
  for (int i = 1; i <= m; i++) {
    int a, b, c, d;
    cin >> a >> b >> c >> d;
    a = f(a);
    b = f(b);
    c = f(c);
    d = f(d);
    if (a > b) {
      swap(a, b);
    }
    if (c > d) {
      swap(c, d);
    }
    ans = 0;
    get(1, 1, n, c, d, a, b);
    cout << ans << "\n";
    ans++;
  }
}