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 Maxn = 1000005;
const int Maxm = 4194304;
int n;
int seqa[Maxn], inb[Maxn];
int in[Maxn];
vector<int> st[Maxm];
int m;
int a, b, c, d;
int x;
void Create(int v, int l, int r) {
  if (l == r)
    st[v].push_back(in[l]);
  else {
    int m = l + r >> 1;
    Create(2 * v, l, m);
    Create(2 * v + 1, m + 1, r);
    int i = 0, j = 0;
    while (i < st[2 * v].size() || j < st[2 * v + 1].size())
      if (i < st[2 * v].size() &&
          (j == st[2 * v + 1].size() || st[2 * v][i] <= st[2 * v + 1][j]))
        st[v].push_back(st[2 * v][i++]);
      else
        st[v].push_back(st[2 * v + 1][j++]);
  }
}
int Get(int v, int l, int r, int a1, int a2, int b1, int b2) {
  int res = 0;
  if (l == a1 && r == a2)
    res = upper_bound(st[v].begin(), st[v].end(), b2) -
          lower_bound(st[v].begin(), st[v].end(), b1);
  else {
    int m = l + r >> 1;
    if (a1 <= m) res += Get(2 * v, l, m, a1, min(m, a2), b1, b2);
    if (m + 1 <= a2)
      res += Get(2 * v + 1, m + 1, r, max(m + 1, a1), a2, b1, b2);
  }
  return res;
}
int f(int z) { return (z - 1 + x + n) % n + 1; }
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &seqa[i]);
  for (int i = 1; i <= n; i++) {
    int b;
    scanf("%d", &b);
    inb[b] = i;
  }
  for (int i = 1; i <= n; i++) in[i] = inb[seqa[i]];
  Create(1, 1, n);
  scanf("%d", &m);
  while (m--) {
    scanf("%d %d %d %d", &a, &b, &c, &d);
    int l1 = min(f(a), f(b)), r1 = max(f(a), f(b)), l2 = min(f(c), f(d)),
        r2 = max(f(c), f(d));
    x = Get(1, 1, n, l1, r1, l2, r2);
    printf("%d\n", x);
    x++;
  }
  return 0;
}