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 DIM = 1e6 + 5;
vector<int> sgt[DIM << 2];
int prm1[DIM], prm2[DIM], pos[DIM];
void build(int n, int l, int r) {
  if (l == r) {
    sgt[n].push_back(pos[prm1[l]]);
    return;
  }
  int m = (l + r) >> 1;
  build(n << 1, l, m);
  build(n << 1 | 1, m + 1, r);
  merge(sgt[n << 1].begin(), sgt[n << 1].end(), sgt[n << 1 | 1].begin(),
        sgt[n << 1 | 1].end(), back_inserter(sgt[n]));
  return;
}
int query(int n, int l, int r, int _l, int _r, int __l, int __r) {
  if (_r < l or r < _l) return 0;
  if (_l <= l and r <= _r)
    return (int)(upper_bound(sgt[n].begin(), sgt[n].end(), __r) -
                 lower_bound(sgt[n].begin(), sgt[n].end(), __l));
  int m = (l + r) >> 1;
  return query(n << 1, l, m, _l, _r, __l, __r) +
         query(n << 1 | 1, m + 1, r, _l, _r, __l, __r);
}
inline int f(int z, int x, int n) { return (z - 1 + x) % n + 1; }
int main(void) {
  int n, x = 0;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &prm1[i]);
  for (int i = 1; i <= n; ++i) scanf("%d", &prm2[i]), pos[prm2[i]] = i;
  int q;
  scanf("%d", &q);
  build(1, 1, n);
  while (q--) {
    int a, b, c, d;
    scanf("%d %d %d %d", &a, &b, &c, &d);
    int l1 = f(a, x, n), r1 = f(b, x, n), l2 = f(c, x, n), r2 = f(d, x, n);
    x = query(1, 1, n, min(l1, r1), max(l1, r1), min(l2, r2), max(l2, r2));
    printf("%d\n", x++);
  }
  return 0;
}