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 M = 1e6 + 5;
int n, a[M];
namespace IST {
int T[M], tot;
int lson[M * 40], rson[M * 40], val[M * 40];
void edit(int O, int& o, int l, int r, int pos) {
  o = ++tot;
  lson[o] = lson[O], rson[o] = rson[O];
  if (l == r) {
    val[o]++;
    return;
  }
  int m = l + r >> 1;
  if (pos <= m)
    edit(lson[O], lson[o], l, m, pos);
  else
    edit(rson[O], rson[o], m + 1, r, pos);
  val[o] = val[lson[o]] + val[rson[o]];
}
int curl(int O, int o, int l, int r, int L, int R) {
  if (L <= l && r <= R) return val[o] - val[O];
  int m = l + r >> 1;
  int t = 0;
  if (L <= m) t += curl(lson[O], lson[o], l, m, L, R);
  if (R > m) t += curl(rson[O], rson[o], m + 1, r, L, R);
  return t;
}
}  // namespace IST
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int x;
    scanf("%d", &x);
    a[x] = i;
  }
  for (int i = 1; i <= n; i++) {
    int x;
    scanf("%d", &x);
    IST::edit(IST::T[i - 1], IST::T[i], 1, n, a[x]);
  }
  int g = 1;
  int q;
  scanf("%d", &q);
  while (q--) {
    int a, b, c, d;
    scanf("%d%d%d%d", &a, &b, &c, &d);
    a = (g + a) % n + 1, b = (g + b) % n + 1, c = (g + c) % n + 1,
    d = (g + d) % n + 1;
    if (a > b) swap(a, b);
    if (c > d) swap(c, d);
    g = IST::curl(IST::T[c - 1], IST::T[d], 1, n, a, b);
    printf("%d\n", g);
  }
}
