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>
int n, m;
int p[1001000], q[1001000], pp[1001000], qq[1001000],
    ans = -1, l1, r1, l2, r2, a, b, c, d, k1, k2, ll1, rr1, ll2, rr2;
int sol[2010][2010];
int main() {
  int i, j;
  scanf("%d", &n);
  for (i = 1; i <= n; ++i) {
    scanf("%d", p + i);
    pp[p[i]] = i;
  }
  for (i = 1; i <= n; ++i) {
    scanf("%d", q + i);
    qq[q[i]] = i;
  }
  for (i = 1; i <= n; ++i) ++sol[i / 500][qq[p[i]] / 500];
  for (i = 0; i <= 2000; ++i)
    for (j = 1; j <= 2000; ++j) sol[i][j] += sol[i][j - 1];
  for (scanf("%d", &m); m; --m) {
    ++ans;
    scanf("%d%d%d%d", &a, &b, &c, &d);
    a += ans;
    if (a > n) a -= n;
    b += ans;
    if (b > n) b -= n;
    c += ans;
    if (c > n) c -= n;
    d += ans;
    if (d > n) d -= n;
    if (a < b) {
      l1 = a;
      r1 = b;
    } else {
      l1 = b;
      r1 = a;
    }
    if (c < d) {
      l2 = c;
      r2 = d;
    } else {
      l2 = d;
      r2 = c;
    }
    ans = 0;
    ll1 = l1 / 500;
    rr1 = r1 / 500;
    ++ll1;
    --rr1;
    ll2 = l2 / 500;
    rr2 = r2 / 500;
    ++ll2;
    --rr2;
    {
      ans = 0;
      for (i = l1; i <= r1; ++i)
        if (l2 <= qq[p[i]] && qq[p[i]] <= r2) ++ans;
      printf("%d\n", ans);
      continue;
    }
    if (r2 - l2 <= 500 * 2) {
      ans = 0;
      for (i = l2; i <= r2; ++i)
        if (l1 <= pp[q[i]] && pp[q[i]] <= r1) ++ans;
      printf("%d\n", ans);
      continue;
    }
    printf("%d\n", ans);
  }
  return 0;
}
