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 = 1000005;
int n, m, tot, x, a[N], b[N], pos[N], rt[N], ls[N * 21], rs[N * 21], sz[N * 21];
int f(int z) { return (z - 1 + x) % n + 1; }
void upd(int &x, int y, int l, int r, int p) {
  x = ++tot, ls[x] = ls[y], rs[x] = rs[y], sz[x] = sz[y] + 1;
  if (l < r) {
    int mid = l + r >> 1;
    if (p <= mid)
      upd(ls[x], ls[y], l, mid, p);
    else
      upd(rs[x], rs[y], mid + 1, r, p);
  }
}
int qry(int x, int l, int r, int ql, int qr) {
  if (!x) return 0;
  if (l >= ql && r <= qr) return sz[x];
  int mid = l + r >> 1, ans = 0;
  if (ql <= mid) ans += qry(ls[x], l, mid, ql, qr);
  if (qr > mid) ans += qry(rs[x], mid + 1, r, ql, qr);
  return ans;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", a + i);
  for (int i = 1; i <= n; i++) scanf("%d", b + i), pos[b[i]] = i;
  for (int i = 1; i <= n; i++) upd(rt[i], rt[i - 1], 1, n, pos[a[i]]);
  scanf("%d", &m);
  for (int a, b, c, d; m--;) {
    scanf("%d%d%d%d", &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);
    printf("%d\n", x = qry(rt[b], 1, n, c, d) - qry(rt[a - 1], 1, n, c, d));
    x++;
  }
}