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 M = 1e6 + 5, Mlog = 21;
int seg[Mlog * M], Rch[Mlog * M], Lch[Mlog * M], root[M], pos[M], n, ans = -1,
                                                                     q, x, cnt;
int f(int z) { return (z + ans) % n + 1; }
int upd(int id, int p, int l = 0, int r = n) {
  int nod = ++cnt;
  if (r - l < 2) return seg[nod] = 1, nod;
  int mid = (l + r) / 2;
  if (p < mid)
    Lch[nod] = upd(Lch[id], p, l, mid), Rch[nod] = Rch[id];
  else
    Rch[nod] = upd(Rch[id], p, mid, r), Lch[nod] = Lch[id];
  return seg[nod] = seg[Lch[nod]] + seg[Rch[nod]], nod;
}
int get(int id, int ql, int qr, int l = 0, int r = n) {
  if (r <= ql || qr <= l) return 0;
  if (ql <= l && r <= qr) return seg[id];
  int mid = (l + r) / 2;
  return get(Lch[id], ql, qr, l, mid) + get(Rch[id], ql, qr, mid, r);
}
int32_t main() {
  cin >> n;
  for (int i = 0; i < n; i++) scanf("%d", &x), pos[x] = i;
  for (int i = 0; i < n; i++)
    scanf("%d", &x), root[i + 1] = upd(root[i], pos[x]);
  cin >> q;
  while (q--) {
    int a, b, c, d, l1, l2, r1, r2;
    scanf("%d%d%d%d", &a, &b, &c, &d), a = f(a), b = f(b), c = f(c), d = f(d);
    l1 = min(a, b), r1 = max(a, b), l2 = min(c, d), r2 = max(c, d);
    printf("%d\n",
           (ans = get(root[r2], l1 - 1, r1) - get(root[l2 - 1], l1 - 1, r1)));
  }
  return 0;
}