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;
int x = 0, n, s;
int f(int z) { return (z - 1 + x) % n + 1; }
const int maxn = 1000005;
int aa[maxn], bb[maxn], pos[maxn], pos2[maxn], bin[maxn];
int tab[1005][1005];
int ret(int b, int d) {
  int r = tab[max(0, bin[b] - 1)][max(0, bin[d] - 1)];
  int k = min(d, (bin[d] - 1) * s);
  for (int i = k + 1; i <= d; i++) {
    if (pos[bb[i]] <= b) r++;
  }
  int k2 = min(b, (bin[b] - 1) * s);
  for (int i = k2 + 1; i <= b; i++) {
    if (pos2[aa[i]] <= k) r++;
  }
  return r;
}
int check(int a, int b, int c, int d) {
  return ret(b, d) - ret(a - 1, d) - ret(b, c - 1) + ret(a - 1, c - 1);
}
int main() {
  scanf("%d", &n);
  s = sqrt(n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &aa[i]);
    pos[aa[i]] = i;
  }
  for (int i = 1; i <= n; i++) bin[i] = (i - 1) / s + 1;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &bb[i]);
    pos2[bb[i]] = i;
    tab[bin[pos[bb[i]]]][bin[i]]++;
  }
  for (int i = 1; i <= bin[n]; i++)
    for (int j = 1; j <= bin[n]; j++) {
      tab[i][j] += tab[i - 1][j] + tab[i][j - 1] - tab[i - 1][j - 1];
    }
  int m;
  scanf("%d", &m);
  while (m--) {
    int a, b, c, d;
    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));
    int ans = check(l1, r1, l2, r2);
    printf("%d\n", ans);
    x = ans + 1;
  }
}
