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 MX_N = (1e6) + 3;
int cord[MX_N][2];
vector<int> tv[MX_N];
int n;
void insert(int x, int y) {
  for (; x <= n; x += (x & (-x))) tv[x].push_back(y);
}
void init() {
  for (int i = 1; i <= n; ++i) sort(tv[i].begin(), tv[i].end());
}
int query(int x, int y) {
  int res = 0;
  for (; x > 0; x -= (x & (-x)))
    res += upper_bound(tv[x].begin(), tv[x].end(), y) - tv[x].begin();
  return res;
}
int main() {
  scanf("%d", &n);
  for (int j = 0; j < 2; ++j)
    for (int i = 1; i <= n; ++i) {
      int x;
      scanf("%d", &x);
      cord[x][j] = i;
    }
  for (int i = 1; i <= n; ++i) insert(cord[i][0], cord[i][1]);
  init();
  int m;
  scanf("%d", &m);
  for (int x = 0; m > 0; --m, ++x) {
    int a[4];
    for (int i = 0; i < 4; ++i) {
      scanf("%d", a + i);
      a[i] = (a[i] - 1 + x) % n + 1;
    }
    int l1 = min(a[0], a[1]);
    int r1 = max(a[0], a[1]);
    int l2 = min(a[2], a[3]);
    int r2 = max(a[2], a[3]);
    x = query(r1, r2) - query(l1 - 1, r2) - query(r1, l2 - 1) +
        query(l1 - 1, l2 - 1);
    printf("%d\n", x);
  }
  return 0;
}