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>
const int N = 1000010, M = 20000010;
int n, m, id, ans, a[N], b[N], c[N], rt[N], ls[M], rs[M], s[M];
inline int f(int x) { return (x - 1 + ans) % n + 1; }
inline void pup(int x) { s[x] = s[ls[x]] + s[rs[x]]; }
int bud(int l, int r) {
  int x = ++id;
  if (l == r) return x;
  ls[x] = bud(l, ((l + r) >> 1));
  rs[x] = bud(((l + r) >> 1) + 1, r);
  return x;
}
int ins(int p, int l, int r, int k) {
  int x = ++id;
  ls[x] = ls[p];
  rs[x] = rs[p];
  s[x] = s[p];
  if (l == r) {
    ++s[x];
    return x;
  }
  if (k <= ((l + r) >> 1))
    ls[x] = ins(ls[p], l, ((l + r) >> 1), k);
  else
    rs[x] = ins(rs[p], ((l + r) >> 1) + 1, r, k);
  pup(x);
  return x;
}
int ask(int lx, int rx, int l, int r, int L, int R) {
  if (l > R || r < L) return 0;
  if (l >= L && r <= R) return s[rx] - s[lx];
  return ask(ls[lx], ls[rx], l, ((l + r) >> 1), L, R) +
         ask(rs[lx], rs[rx], ((l + r) >> 1) + 1, r, L, R);
}
int main() {
  scanf("%d", &n);
  rt[0] = bud(1, n);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &b[i]);
    c[b[i]] = i;
  }
  for (int i = 1; i <= n; ++i) rt[i] = ins(rt[i - 1], 1, n, c[a[i]]);
  scanf("%d", &m);
  for (int a, b, c, d, l1, r1, l2, r2, i = 1; i <= m; ++i) {
    scanf("%d%d%d%d", &a, &b, &c, &d);
    a = f(a);
    b = f(b);
    c = f(c);
    d = f(d);
    l1 = std::min(a, b);
    r1 = std::max(a, b);
    l2 = std::min(c, d);
    r2 = std::max(c, d);
    printf("%d\n", ans = ask(rt[l2 - 1], rt[r2], 1, n, l1, r1));
    ++ans;
  }
}
