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 = 1e6 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 7;
int n, i, x, m, a, b, c, d, l, r, l2, r2, lastans;
int H[N], A[N];
vector<int> first[4 * N];
int f(int x) { return (x - 1 + lastans) % n + 1; }
int query(int root, int bas, int son, int x, int y, int x2, int y2) {
  if (bas > y or son < x) return 0;
  if (x <= bas and son <= y)
    return upper_bound(first[root].begin(), first[root].end(), y2) -
           lower_bound(first[root].begin(), first[root].end(), x2);
  return query((root + root), bas, ((bas + son) / 2), x, y, x2, y2) +
         query((root + root + 1), ((bas + son) / 2) + 1, son, x, y, x2, y2);
}
void init(int root, int bas, int son) {
  if (bas == son) {
    first[root].push_back(A[bas]);
    return;
  }
  init((root + root), bas, ((bas + son) / 2));
  init((root + root + 1), ((bas + son) / 2) + 1, son);
  first[root].resize(first[(root + root)].size() +
                     first[(root + root + 1)].size());
  merge(first[(root + root)].begin(), first[(root + root)].end(),
        first[(root + root + 1)].begin(), first[(root + root + 1)].end(),
        first[root].begin());
}
int main() {
  scanf("%d", &n);
  for (i = 1; i <= n; i++) scanf("%d", A + i);
  for (i = 1; i <= n; i++) {
    scanf("%d", &x);
    H[x] = i;
  }
  for (i = 1; i <= n; i++) A[i] = H[A[i]];
  init(1, 1, n);
  scanf("%d", &m);
  for (i = 1; i <= m; i++) {
    scanf("%d %d %d %d", &a, &b, &c, &d);
    l = min(f(a), f(b));
    r = max(f(a), f(b));
    l2 = min(f(c), f(d));
    r2 = max(f(c), f(d));
    lastans = query(1, 1, n, l, r, l2, r2) + 1;
    printf("%d\n", lastans - 1);
  }
}