Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int in() {
  int32_t x;
  scanf("%d", &x);
  return x;
}
inline long long lin() {
  long long x;
  scanf("%lld", &x);
  return x;
}
const int maxn = 1e6 + 10;
int n;
int x = 0;
pair<int, int> where[maxn];
inline int f(int z) { return (z - 1 + x) % n; }
int stree[21][maxn];
inline void build(int p = 0, int l = 0, int r = n - 1, int h = 0) {
  if (l == r) {
    stree[h][l] = where[l].second;
    return;
  }
  int mid = (l + r) >> 1;
  build(2 * p + 1, l, mid, h + 1), build(2 * p + 2, mid + 1, r, h + 1);
  merge(stree[h + 1] + l, stree[h + 1] + mid + 1, stree[h + 1] + mid + 1,
        stree[h + 1] + r + 1, stree[h] + l);
}
inline int get(int ql, int qr, int ql2, int qr2, int p = 0, int l = 0,
               int r = n - 1, int h = 0) {
  if (l > qr || r < ql || ql > qr) return 0;
  if (l >= ql && r <= qr) {
    return upper_bound(stree[h] + l, stree[h] + r + 1, qr2) -
           lower_bound(stree[h] + l, stree[h] + r + 1, ql2);
  }
  int mid = (l + r) >> 1;
  return get(ql, qr, ql2, qr2, 2 * p + 1, l, mid, h + 1) +
         get(ql, qr, ql2, qr2, 2 * p + 2, mid + 1, r, h + 1);
}
int32_t main() {
  n = in();
  for (int i = 0; i < n; i++) {
    int foo = in() - 1;
    where[foo].first = i;
  }
  for (int i = 0; i < n; i++) {
    int foo = in() - 1;
    where[foo].second = i;
  }
  sort(where, where + n);
  build();
  int q = in();
  while (q--) {
    int a = in(), b = in(), c = in(), d = in();
    int l1 = min(f(a), f(b)), r1 = max(f(a), f(b)), l2 = min(f(c), f(d)),
        r2 = max(f(c), f(d));
    int ret = get(l1, r1, l2, r2);
    cout << ret << "\n";
    x = ret + 1;
  }
}