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;
const int N = (int)1e6 + 10;
int n, a[N], pos[N];
struct SegTree {
  vector<int> S[N * 5];
  void Build(int k, int l, int r) {
    if (l == r) {
      S[k].push_back(a[l]);
      return;
    }
    Build(k << 1 | 0, l, ((l + r) >> 1));
    Build(k << 1 | 1, ((l + r) >> 1) + 1, r);
    S[k].assign(r - l + 1, 0);
    merge(S[k << 1 | 0].begin(), S[k << 1 | 0].end(), S[k << 1 | 1].begin(),
          S[k << 1 | 1].end(), S[k].begin());
  }
  int Count(int k, int l, int r) {
    vector<int>::iterator i1 = lower_bound(S[k].begin(), S[k].end(), l);
    vector<int>::iterator i2 = upper_bound(S[k].begin(), S[k].end(), r);
    return i2 - i1;
  }
  int Query(int k, int l, int r, int i, int j, int x, int y) {
    if (l > j || r < i) return 0;
    if (i <= l && r <= j) return Count(k, x, y);
    return Query(k << 1 | 0, l, ((l + r) >> 1), i, j, x, y) +
           Query(k << 1 | 1, ((l + r) >> 1) + 1, r, i, j, x, y);
  }
} ST;
int f(int z, int x) { return (z - 1 + x) % n + 1; }
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  for (int i = 1; i <= n; ++i) {
    int x;
    scanf("%d", &x);
    pos[x] = i;
  }
  for (int i = 1; i <= n; ++i) a[i] = pos[a[i]];
  ST.Build(1, 1, n);
  int q;
  scanf("%d", &q);
  int lastAns = -1;
  while (q--) {
    lastAns++;
    int a, b, l1, r1;
    scanf("%d%d", &a, &b);
    int c, d, l2, r2;
    scanf("%d%d", &c, &d);
    int x, y;
    x = f(a, lastAns);
    y = f(b, lastAns);
    l1 = min(x, y);
    r1 = max(x, y);
    x = f(c, lastAns);
    y = f(d, lastAns);
    l2 = min(x, y);
    r2 = max(x, y);
    printf("%d\n", lastAns = ST.Query(1, 1, n, l1, r1, l2, r2));
  }
  return 0;
}