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 read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int n, m, a[1000100], b[1000100], pos[1000100], f[1000100], l1, l2, r1, r2, ans;
int tr[1000100 * 10], root[1000100], tot = 0, s[1000100 * 10], lt[1000100 * 10],
                                     rt[1000100 * 10];
void build(int &x, int l, int r) {
  x = ++tot;
  if (l == r) {
    s[x] = 0;
    return;
  }
  int mid = (l + r) >> 1;
  build(lt[x], l, mid);
  build(rt[x], mid + 1, r);
}
void ins(int &x, int l, int r, int last, int k) {
  x = ++tot;
  s[x] = s[last];
  s[x]++;
  lt[x] = lt[last];
  rt[x] = rt[last];
  if (l == r) return;
  int mid = (l + r) >> 1;
  if (k <= mid)
    ins(lt[x], l, mid, lt[last], k);
  else
    ins(rt[x], mid + 1, r, rt[last], k);
}
int ask(int x, int l, int r, int L, int R) {
  if (l == L && r == R) return s[x];
  int mid = (l + r) >> 1;
  if (R <= mid)
    return ask(lt[x], l, mid, L, R);
  else if (L > mid)
    return ask(rt[x], mid + 1, r, L, R);
  return ask(lt[x], l, mid, L, mid) + ask(rt[x], mid + 1, r, mid + 1, R);
}
int fu(int x) { return ((x - 1 + ans) % n) + 1; }
int main() {
  n = read();
  for (int i = 1; i <= n; i++) a[i] = read();
  for (int i = 1; i <= n; i++) b[i] = read(), pos[b[i]] = i;
  for (int i = 1; i <= n; i++) f[i] = pos[a[i]];
  build(root[0], 1, n);
  for (int i = 1; i <= n; i++) {
    ins(root[i], 1, n, root[i - 1], f[i]);
  }
  m = read();
  ans = 0;
  while (m--) {
    int x, y, z, w;
    x = read();
    y = read();
    z = read();
    w = read();
    x = fu(x);
    y = fu(y);
    z = fu(z);
    w = fu(w);
    if (x > y)
      r1 = x, l1 = y;
    else
      l1 = x, r1 = y;
    if (z > w)
      r2 = z, l2 = w;
    else
      l2 = z, r2 = w;
    ans = ask(root[r1], 1, n, l2, r2) - ask(root[l1 - 1], 1, n, l2, r2);
    printf("%d\n", ans);
    ans++;
  }
  return 0;
}
