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>
class CcT {
 public:
  int l[21000200], r[21000200], a[21000200];
  int n;
  int ins(int o, int p, int s) {
    int m = ++n;
    a[m] = a[o] + 1;
    if (s == 1) return m;
    if (p < (s >> 1))
      l[m] = ins(l[o], p, s >> 1), r[m] = r[o];
    else
      l[m] = l[o], r[m] = ins(r[o], p - (s >> 1), s >> 1);
    return m;
  }
  int find(int o, int p, int s) {
    if (s == 1)
      return a[o];
    else if (p < (s >> 1))
      return find(l[o], p, s >> 1);
    else
      return find(r[o], p - (s >> 1), s >> 1) + a[l[o]];
  }
} X;
void gi(int &x) {
  char ch = getchar();
  x = 0;
  while (ch < '0' || ch > '9') ch = getchar();
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - 48, ch = getchar();
}
void pi(int x) {
  if (x > 9) pi(x / 10);
  putchar(x % 10 + 48);
}
int p[1000020], q[1000020], P[1000020], s[1000020], root[1000020];
int n, m, x, l1, r1, l2, r2, a, b, c, d, i, j, k;
int main() {
  gi(n);
  for (i = 1; i <= n; i++) gi(p[i]);
  for (i = 1; i <= n; i++) gi(q[i]);
  for (i = 1; i <= n; i++) P[q[i]] = i;
  for (i = 1; i <= n; i++) s[i] = P[p[i]];
  for (i = 1; i <= n; i++) root[i] = X.ins(root[i - 1], s[i], 1048576);
  gi(m);
  x = 0;
  while (m--) {
    gi(a);
    gi(b);
    gi(c);
    gi(d);
    l1 = (((a - 1 + x) % n) + 1);
    r1 = (((b - 1 + x) % n) + 1);
    l2 = (((c - 1 + x) % n) + 1);
    r2 = (((d - 1 + x) % n) + 1);
    if (l1 > r1) l1 ^= r1, r1 ^= l1, l1 ^= r1;
    if (l2 > r2) l2 ^= r2, r2 ^= l2, l2 ^= r2;
    l1--;
    l2--;
    x = X.find(root[l1], l2, 1048576) + X.find(root[r1], r2, 1048576) -
        X.find(root[l1], r2, 1048576) - X.find(root[r1], l2, 1048576);
    pi(x);
    putchar('\n');
    x++;
  }
  return 0;
}