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;
const int Maxn = 1000005;
struct node {
  node *pnt[4];
  int has;
  node() {
    pnt[0] = pnt[1] = pnt[2] = pnt[3] = NULL;
    has = 0;
  }
};
int n;
int ina[Maxn], inb[Maxn];
node qt;
int m;
int a, b, c, d;
int x;
void Insert(node &n, int l1, int r1, int l2, int r2, int a, int b) {
  n.has++;
  if (l1 < r1 || l2 < r2) {
    int m1 = l1 + r1 >> 1, m2 = l2 + r2 >> 1;
    if (a <= m1 && b <= m2) {
      if (n.pnt[0] == NULL) n.pnt[0] = new node();
      Insert(*n.pnt[0], l1, m1, l2, m2, a, b);
    }
    if (a > m1 && b <= m2) {
      if (n.pnt[1] == NULL) n.pnt[1] = new node();
      Insert(*n.pnt[1], m1 + 1, r1, l2, m2, a, b);
    }
    if (a <= m1 && b > m2) {
      if (n.pnt[2] == NULL) n.pnt[2] = new node();
      Insert(*n.pnt[2], l1, m1, m2 + 1, r2, a, b);
    }
    if (a > m1 && b > m2) {
      if (n.pnt[3] == NULL) n.pnt[3] = new node();
      Insert(*n.pnt[3], m1 + 1, r1, m2 + 1, r2, a, b);
    }
  }
}
int Get(node &n, int l1, int r1, int l2, int r2, int a1, int a2, int b1,
        int b2) {
  int res = 0;
  if (l1 == a1 && r1 == a2 && l2 == b1 && r2 == b2)
    res = n.has;
  else {
    int m1 = l1 + r1 >> 1, m2 = l2 + r2 >> 1;
    if (a1 <= m1 && b1 <= m2 && n.pnt[0] != NULL)
      res += Get(*n.pnt[0], l1, m1, l2, m2, a1, min(m1, a2), b1, min(m2, b2));
    if (a2 > m1 && b1 <= m2 && n.pnt[1] != NULL)
      res += Get(*n.pnt[1], m1 + 1, r1, l2, m2, max(m1 + 1, a1), a2, b1,
                 min(m2, b2));
    if (a1 <= m2 && b2 > m2 && n.pnt[2] != NULL)
      res += Get(*n.pnt[2], l1, m1, m2 + 1, r2, a1, min(m1, a2),
                 max(m2 + 1, b1), b2);
    if (a2 > m1 && b2 > m2 && n.pnt[3] != NULL)
      res += Get(*n.pnt[3], m1 + 1, r1, m2 + 1, r2, max(m1 + 1, a1), a2,
                 max(m2 + 1, b1), b2);
  }
  return res;
}
int f(int z) { return (z - 1 + x + n) % n + 1; }
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int a;
    scanf("%d", &a);
    ina[a] = i;
  }
  for (int i = 1; i <= n; i++) {
    int b;
    scanf("%d", &b);
    inb[b] = i;
  }
  for (int i = 1; i <= n; i++) Insert(qt, 1, n, 1, n, ina[i], inb[i]);
  scanf("%d", &m);
  while (m--) {
    scanf("%d %d %d %d", &a, &b, &c, &d);
    int l1 = min(f(a), f(b)), r1 = max(f(a), f(b)), l2 = min(f(c), f(d)),
        r2 = max(f(c), f(d));
    x = Get(qt, 1, n, 1, n, l1, r1, l2, r2);
    printf("%d\n", x);
    x++;
  }
  return 0;
}
