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;
const int oo = 1047483647, maxn = 1001000;
int n, i, j, k, m, q, tot, ans;
int f[maxn * 20], a[maxn], L[maxn * 20], R[maxn * 20], b[maxn], root[maxn],
    w[maxn];
int Read() {
  char ch;
  while (ch = getchar(), (ch < '0' || ch > '9') && (ch != '-'))
    ;
  bool neg = (ch == '-');
  if (ch == '-') ch = getchar();
  int v = 0;
  while (ch >= '0' && ch <= '9') v = v * 10 + ch - '0', ch = getchar();
  if (neg) v = -v;
  return v;
}
void xiu(int &t, int l, int r, int w, int now) {
  t = (++tot);
  f[t] = f[now] + 1;
  if (l == r) return;
  int mid = (l + r) >> 1;
  if (w <= mid) {
    R[t] = R[now];
    xiu(L[t], l, mid, w, L[now]);
  } else {
    L[t] = L[now];
    xiu(R[t], mid + 1, r, w, R[now]);
  }
}
int change(int x) { return (x - 1 + ans) % n + 1; }
int ask(int t, int x, int y, int l, int r) {
  if (!t) return 0;
  if (l <= x && y <= r) {
    return f[t];
  }
  int mid = (x + y) >> 1;
  if (r <= mid)
    return ask(L[t], x, mid, l, r);
  else if (l > mid)
    return ask(R[t], mid + 1, y, l, r);
  else {
    return ask(L[t], x, mid, l, r) + ask(R[t], mid + 1, y, l, r);
  }
}
int main() {
  n = Read();
  for (i = 1; i <= n; i++) a[i] = Read();
  for (i = 1; i <= n; i++) b[i] = Read(), w[b[i]] = i;
  for (i = 1; i <= n; i++) {
    int v = w[a[i]];
    xiu(root[i], 1, n, v, root[i - 1]);
  }
  m = Read();
  for (; m; m--) {
    int l1, l2, r1, r2;
    l1 = Read();
    r1 = Read();
    l2 = Read();
    r2 = Read();
    l1 = change(l1);
    l2 = change(l2);
    r1 = change(r1);
    r2 = change(r2);
    if (l1 > r1) swap(l1, r1);
    if (l2 > r2) swap(l2, r2);
    ans = ask(root[r1], 1, n, l2, r2) - ask(root[l1 - 1], 1, n, l2, r2);
    printf("%d\n", ans);
    ans++;
  }
  return 0;
}
