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;
vector<int> t[1000010 << 2];
int a[1000010], b[1000010];
void ins(int p, int l, int r) {
  if (l == r) {
    t[p].resize(1);
    t[p][0] = b[a[l]];
    return;
  }
  int m = (l + r) >> 1;
  ins(p << 1, l, m);
  ins(p << 1 | 1, m + 1, r);
  t[p].resize(t[p << 1].size() + t[p << 1 | 1].size());
  merge(t[p << 1].begin(), t[p << 1].end(), t[p << 1 | 1].begin(),
        t[p << 1 | 1].end(), t[p].begin());
}
int get(int p, int l, int r, int first, int second, int a, int b) {
  if (second < l || first > r) return 0;
  if (l >= first && r <= second) {
    return upper_bound(t[p].begin(), t[p].end(), b) -
           lower_bound(t[p].begin(), t[p].end(), a);
  }
  int m = (l + r) >> 1;
  return get(p << 1, l, m, first, second, a, b) +
         get(p << 1 | 1, m + 1, r, first, second, a, b);
}
int main() {
  int i, j, n, m;
  while (~scanf("%d", &n)) {
    for (i = 1; i <= n; i++) scanf("%d", &a[i]);
    for (i = 1; i <= n; i++) scanf("%d", &j), b[j] = i;
    ins(1, 1, n);
    scanf("%d", &m);
    int first = 0;
    while (m--) {
      int c, d, e, f;
      scanf("%d%d%d%d", &c, &d, &e, &f);
      c = (c - 1 + first) % n + 1;
      d = (d - 1 + first) % n + 1;
      e = (e - 1 + first) % n + 1;
      f = (f - 1 + first) % n + 1;
      int l1 = min(c, d), r1 = max(c, d);
      int l2 = min(e, f), r2 = max(e, f);
      first = get(1, 1, n, l1, r1, l2, r2);
      printf("%d\n", first);
      first++;
    }
  }
  return 0;
}