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;
int n, p[1000005], q[1000005];
vector<int> rt[2000005];
void init() {
  for (int i = (n); i < (n + n); i++) rt[i].push_back(q[i - n]);
  for (int i = n - 1; i; i--)
    merge(rt[2 * i].begin(), rt[2 * i].end(), rt[2 * i + 1].begin(),
          rt[2 * i + 1].end(), back_inserter(rt[i]));
}
int qu(int l, int r, int a, int b) {
  int res = 0;
  for (l += n, r += n; l < r; l /= 2, r /= 2) {
    if (l & 1) {
      res += lower_bound(rt[l].begin(), rt[l].end(), b) -
             lower_bound(rt[l].begin(), rt[l].end(), a);
      l++;
    }
    if (r & 1) {
      r--;
      res += lower_bound(rt[r].begin(), rt[r].end(), b) -
             lower_bound(rt[r].begin(), rt[r].end(), a);
    }
  }
  return res;
}
int main() {
  scanf("%d", &n);
  for (int i = (0); i < (n); i++) {
    int x;
    scanf("%d", &x);
    p[x - 1] = i;
  }
  for (int i = (0); i < (n); i++) {
    int x;
    scanf("%d", &x);
    q[i] = p[x - 1];
  }
  init();
  int m;
  scanf("%d", &m);
  int x = 0;
  for (int i = (0); i < (m); i++) {
    int a, b, c, d;
    scanf("%d%d%d%d", &a, &b, &c, &d);
    a = ((a - 1 + x) % n), b = ((b - 1 + x) % n), c = ((c - 1 + x) % n),
    d = ((d - 1 + x) % n);
    if (a > b) swap(a, b);
    if (c > d) swap(c, d);
    b++;
    d++;
    x = qu(c, d, a, b) + 1;
    printf("%d\n", x - 1);
  }
}