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 n, T;
int tot, cnt, it, lastans;
int a[1000010], b[1000010], rx[1000010], ly[1000010 << 5], ry[1000010 << 5],
    sum[1000010 << 5];
void build(int &s, int l, int r) {
  s = ++tot;
  if (l == r) return;
  int mid = (l + r) >> 1;
  build(ly[s], l, mid);
  build(ry[s], mid + 1, r);
}
int update(int x, int l, int r, int p) {
  int y = ++tot;
  ly[y] = ly[x];
  ry[y] = ry[x];
  sum[y] = sum[x] + 1;
  if (l == r) return y;
  int mid = (l + r) >> 1;
  if (p <= mid)
    ly[y] = update(ly[y], l, mid, p);
  else
    ry[y] = update(ry[y], mid + 1, r, p);
  return y;
}
int query(int p, int l, int r, int L, int R) {
  int ans;
  if (l == L && r == R) return sum[p];
  int mid = (l + r) >> 1;
  if (R <= mid)
    return query(ly[p], l, mid, L, R);
  else if (L > mid)
    return query(ry[p], mid + 1, r, L, R);
  else
    return query(ly[p], l, mid, L, mid) + query(ry[p], mid + 1, r, mid + 1, R);
}
int f(int x) { return (x - 1 + lastans) % n + 1; }
int main() {
  scanf("%d", &n);
  for (int i = 1, x; i <= n; i++) {
    scanf("%d", &x);
    b[x] = i;
  }
  for (int i = 1, x; i <= n; i++) {
    scanf("%d", &x);
    a[i] = b[x];
  }
  build(rx[0], 1, n);
  for (int i = 1; i <= n; i++) {
    rx[i] = update(rx[i - 1], 1, n, a[i]);
  }
  scanf("%d", &T);
  while (T--) {
    int l1, l2, r1, r2;
    scanf("%d %d %d %d", &l1, &r1, &l2, &r2);
    l1 = f(l1), r1 = f(r1), l2 = f(l2), r2 = f(r2);
    if (l1 > r1) swap(l1, r1);
    if (l2 > r2) swap(l2, r2);
    int ans = query(rx[r2], 1, n, l1, r1) - query(rx[l2 - 1], 1, n, l1, r1);
    printf("%d\n", ans);
    lastans++;
  }
  return 0;
}
