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;
const int M = N * 22;
int tsz, root[N], ls[M], rs[M], sum[M];
void Set(int p, int &c, int ss, int se, int qi) {
  c = ++tsz;
  ls[c] = ls[p];
  rs[c] = rs[p];
  sum[c] = sum[p] + 1;
  if (ss == se) return;
  int mid = ss + se >> 1;
  if (qi <= mid)
    Set(ls[p], ls[c], ss, mid, qi);
  else
    Set(rs[p], rs[c], mid + 1, se, qi);
}
int Get(int c, int ss, int se, int qs, int qe) {
  if (qs > qe || qs > se || ss > qe) return 0;
  if (qs <= ss && qe >= se) return sum[c];
  int mid = ss + se >> 1;
  return Get(ls[c], ss, mid, qs, qe) + Get(rs[c], mid + 1, se, qs, qe);
}
int x[N], y[N], p[N], q[N];
int main() {
  int n, qu;
  scanf("%i", &n);
  for (int i = 1; i <= n; i++) scanf("%i", &p[i]), x[p[i]] = i;
  for (int i = 1; i <= n; i++) scanf("%i", &q[i]), y[q[i]] = i;
  for (int i = 1; i <= n; i++) Set(root[i - 1], root[i], 1, n, y[p[i]]);
  int ans = 0;
  auto f = [&](int x) { return (x - 1 + ans) % n + 1; };
  scanf("%i", &qu);
  while (qu--) {
    int a, b, c, d;
    scanf("%i %i %i %i", &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 = Get(root[b], 1, n, c, d) - Get(root[a - 1], 1, n, c, d);
    printf("%i\n", ans);
    ans++;
  }
  return 0;
}