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 NMAX = 4400000;
int x, n;
int f(int z) { return ((z - 1 + x) % n); }
int a[NMAX], b[NMAX], b1[NMAX], l1, r1, l2, r2;
vector<int> md[NMAX];
void build(int idx, int l, int r) {
  int len = r - l + 1;
  md[idx].resize(len);
  if (l == r) {
    md[idx][0] = a[l];
    return;
  }
  int mid = (l + r) >> 1;
  build(idx << 1, l, mid);
  build(idx << 1 | 1, mid + 1, r);
  merge(md[idx << 1].begin(), md[idx << 1].end(), md[idx << 1 | 1].begin(),
        md[idx << 1 | 1].end(), md[idx].begin());
}
int query(int idx, int l, int r) {
  if (r < l1 || r1 < l) return 0;
  if (l1 <= l && r <= r1) {
    if (r2 < md[idx][0] || md[idx].back() < l2) return 0;
    return int(upper_bound(md[idx].begin(), md[idx].end(), r2) -
               lower_bound(md[idx].begin(), md[idx].end(), l2));
  }
  int mid = (l + r) >> 1;
  return query(idx << 1, l, mid) + query(idx << 1 | 1, mid + 1, r);
}
int main() {
  cin >> n;
  for (int i = 0; i < (int)(n); i++) {
    scanf("%d", &a[i]);
    a[i]--;
  }
  for (int i = 0; i < (int)(n); i++) {
    scanf("%d", &b[i]);
    b[i]--;
  }
  for (int i = 0; i < (int)(n); i++) b1[b[i]] = i;
  for (int i = 0; i < (int)(n); i++) a[i] = b1[a[i]];
  build(1, 0, n - 1);
  int m;
  scanf("%d", &m);
  for (int i = 0; i < (int)(m); i++) {
    int a, b, c, d;
    scanf("%d%d%d%d", &a, &b, &c, &d);
    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 = query(1, 0, n - 1);
    printf("%d\n", x);
    x++;
  }
  return 0;
}