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 n, m, l1, l2, r1, r2, t, a[1000005], b[1000005], tot, ans;
struct node {
  int lson, rson, k, q;
} f[4000005];
void add(int &kk, int l, int r, int x, int ff) {
  tot++;
  kk = tot;
  f[kk].lson = f[ff].lson;
  f[kk].rson = f[ff].rson;
  f[kk].k = f[ff].k + 1;
  if (l == r) return;
  int mid = (l + r) >> 1;
  if (x <= mid)
    add(f[kk].lson, l, mid, x, f[ff].lson);
  else
    add(f[kk].rson, mid + 1, r, x, f[ff].rson);
  return;
}
int solve(int kk, int l, int r, int ll, int rr) {
  if (ll <= l && rr >= r) return f[kk].k;
  int mid = (l + r) >> 1;
  t = 0;
  if (ll <= mid) t = solve(f[kk].lson, l, mid, ll, rr);
  if (rr > mid) t += solve(f[kk].rson, mid + 1, r, ll, rr);
  return t;
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> t;
    a[t] = i;
  }
  for (int i = 1; i <= n; i++) {
    cin >> t;
    b[i] = a[t];
  }
  for (int i = 1; i <= n; i++) {
    add(f[i].q, 1, n, b[i], f[i - 1].q);
  }
  cin >> m;
  while (m--) {
    cin >> l1 >> r1 >> l2 >> r2;
    l1 = (l1 - 1 + ans) % n + 1;
    r1 = (r1 - 1 + ans) % n + 1;
    l2 = (l2 - 1 + ans) % n + 1;
    r2 = (r2 - 1 + ans) % n + 1;
    if (l1 > r1) swap(l1, r1);
    if (l2 > r2) swap(l2, r2);
    ans = solve(f[r2].q, 1, n, l1, r1) - solve(f[l2 - 1].q, 1, n, l1, r1);
    cout << ans << endl;
    ans++;
  }
  return 0;
}
