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 M = 1 << 20;
const int N = 1e6 + 100;
int n, X;
vector<int> stree[M * 2];
int p[N], q[N];
int qu(int l1, int r1, int l2, int r2, int small = 0, int big = M - 1,
       int index = 1) {
  l1 = max(l1, small);
  r1 = min(r1, big);
  if (l1 > r1) return 0;
  if (small == l1 and big == r1)
    return upper_bound((stree[index]).begin(), (stree[index]).end(), r2) -
           lower_bound((stree[index]).begin(), (stree[index]).end(), l2);
  return qu(l1, r1, l2, r2, small, (small + big) / 2, 2 * index) +
         qu(l1, r1, l2, r2, (small + big) / 2 + 1, big, index * 2 + 1);
}
int main() {
  ios_base::sync_with_stdio(0);
  int tmp;
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> tmp;
    p[tmp] = i;
  }
  for (int i = 1; i <= n; ++i) {
    cin >> tmp;
    q[tmp] = i;
  }
  for (int i = 1; i <= n; ++i) stree[p[i] + M].push_back(q[i]);
  for (int i = M - 1; i >= 0; --i)
    merge((stree[i * 2]).begin(), (stree[i * 2]).end(),
          (stree[i * 2 + 1]).begin(), (stree[i * 2 + 1]).end(),
          back_inserter(stree[i]));
  int m, a, b, c, d;
  cin >> m;
  while (m--) {
    cin >> a >> b >> c >> d;
    int l1 = min(((((a)-1 + X) % n) + 1), ((((b)-1 + X) % n) + 1)),
        r1 = max(((((a)-1 + X) % n) + 1), ((((b)-1 + X) % n) + 1)),
        l2 = min(((((c)-1 + X) % n) + 1), ((((d)-1 + X) % n) + 1)),
        r2 = max(((((c)-1 + X) % n) + 1), ((((d)-1 + X) % n) + 1));
    X = qu(l1, r1, l2, r2);
    cout << X++ << endl;
  }
  return 0;
}