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;
static const int INF = 500000000;
template <class T>
void debug(T a, T b) {
  for (; a != b; ++a) cerr << *a << ' ';
  cerr << endl;
}
struct segtree {
  vector<int> val[2200000];
  int n;
  void init(int n_, int *ar) {
    n = 1;
    while (n < n_) n <<= 1;
    for (int i = 0; i < n_; ++i) {
      val[i + n - 1].push_back(ar[i]);
    }
    for (int i = n - 2; i >= 0; --i) {
      val[i].resize(val[i * 2 + 1].size() + val[i * 2 + 2].size());
      merge((val[i * 2 + 1]).begin(), (val[i * 2 + 1]).end(),
            (val[i * 2 + 2]).begin(), (val[i * 2 + 2]).end(), val[i].begin());
    }
  }
  int query(int a, int b, int i, int l, int r, int ub, int lb) {
    if (a <= l && r <= b) {
      return lower_bound((val[i]).begin(), (val[i]).end(), ub) -
             lower_bound((val[i]).begin(), (val[i]).end(), lb);
    }
    if (b <= l || r <= a) return 0;
    int md = (l + r) >> 1;
    return query(a, b, i * 2 + 1, l, md, ub, lb) +
           query(a, b, i * 2 + 2, md, r, ub, lb);
  }
};
segtree seg;
int n;
int perm[1000005], perm2[1000005];
int pos2[1000005], crsp[1000005];
int x = 0;
int f(int a) { return (a - 1 + x) % n; }
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &perm[i]);
    --perm[i];
  }
  for (int i = 0; i < n; ++i) {
    scanf("%d", &perm2[i]);
    --perm2[i];
    pos2[perm2[i]] = i;
  }
  for (int i = 0; i < n; ++i) crsp[i] = pos2[perm[i]];
  seg.init(n, crsp);
  int m;
  scanf("%d", &m);
  for (int hoge = 0; hoge < m; ++hoge) {
    int a, b, c, d;
    scanf("%d%d%d%d", &a, &b, &c, &d);
    int A = f(a), B = f(b), C = f(c), D = f(d);
    int l1 = min(A, B), r1 = max(A, B), l2 = min(C, D), r2 = max(C, D);
    ++r1;
    ++r2;
    int res = seg.query(l1, r1, 0, 0, seg.n, r2, l2);
    x = res + 1;
    printf("%d\n", res);
  }
  return 0;
}