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 MAXT = 20000005;
int top;
int s[MAXT], L[MAXT], R[MAXT];
int Query(int k, int l, int r, int x, int y) {
  if (x == l && y == r) return s[k];
  int m = l + r >> 1;
  if (y <= m) return Query(L[k], l, m, x, y);
  if (x > m) return Query(R[k], m + 1, r, x, y);
  return Query(L[k], l, m, x, m) + Query(R[k], m + 1, r, m + 1, y);
}
int Insert(int k, int l, int r, int d) {
  int t = ++top;
  s[t] = s[k] + 1, L[t] = L[k], R[t] = R[k];
  if (l == r) return t;
  int m = l + r >> 1;
  if (d <= m) return L[t] = Insert(L[k], l, m, d), t;
  return R[t] = Insert(R[k], m + 1, r, d), t;
}
int d[1000005], root[1000005];
int main() {
  int n, m;
  scanf("%d", &n);
  for (int i = 1, x; i <= n; i++) scanf("%d", &x), d[x] = i;
  for (int i = 1, x; i <= n; i++)
    scanf("%d", &x), root[i] = Insert(root[i - 1], 1, n, d[x]);
  scanf("%d", &m);
  for (int a, b, x, y, lasans = 0; m--;) {
    scanf("%d%d%d%d", &a, &b, &x, &y);
    a = ((a - 1 + lasans) % n) + 1;
    b = ((b - 1 + lasans) % n) + 1;
    x = ((x - 1 + lasans) % n) + 1;
    y = ((y - 1 + lasans) % n) + 1;
    if (a > b) swap(a, b);
    if (x > y) swap(x, y);
    lasans = Query(root[b], 1, n, x, y) - Query(root[a - 1], 1, n, x, y);
    printf("%d\n", lasans++);
  }
}
