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 long long mod = 1e9 + 7;
const int N = 1e6 + 5;
int n, x = 0;
int a[N], b[N], arr[N], pos[N];
struct {
  vector<int> v;
} t[4 * N];
int fun(int id) { return (id - 1 + x) % n + 1; }
void construct(int node, int start, int end) {
  if (start == end)
    t[node].v.push_back(pos[start]);
  else {
    int mid = (start + end) >> 1;
    construct(2 * node, start, mid);
    construct(2 * node + 1, mid + 1, end);
    int i = 0, j = 0, n1 = t[2 * node].v.size(), n2 = t[2 * node + 1].v.size();
    while (i < n1 && j < n2) {
      if (t[2 * node].v[i] < t[2 * node + 1].v[j]) {
        t[node].v.push_back(t[2 * node].v[i]);
        i++;
      } else {
        t[node].v.push_back(t[2 * node + 1].v[j]);
        j++;
      }
    }
    while (i < n1) {
      t[node].v.push_back(t[2 * node].v[i]);
      i++;
    }
    while (j < n2) {
      t[node].v.push_back(t[2 * node + 1].v[j]);
      j++;
    }
  }
}
int query(int node, int start, int end, int l, int r, int x, int y) {
  if (start > end || start > r || end < l) return 0;
  if (start >= l && end <= r) {
    int id1, id2;
    id1 =
        lower_bound(t[node].v.begin(), t[node].v.end(), x) - t[node].v.begin();
    id2 = upper_bound(t[node].v.begin(), t[node].v.end(), y) -
          t[node].v.begin() - 1;
    if (id1 <= id2) return id2 - id1 + 1;
    return 0;
  }
  int mid = (start + end) >> 1;
  int q1 = query(2 * node, start, mid, l, r, x, y);
  int q2 = query(2 * node + 1, mid + 1, end, l, r, x, y);
  return q1 + q2;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  ;
  cin >> n;
  int i;
  for (i = 1; i <= n; i++) cin >> a[i];
  for (i = 1; i <= n; i++) {
    cin >> b[i];
    arr[b[i]] = i;
  }
  for (i = 1; i <= n; i++) pos[i] = arr[a[i]];
  construct(1, 1, n);
  int m;
  cin >> m;
  while (m--) {
    int p, q, r, s;
    cin >> p >> q >> r >> s;
    int l1 = fun(p), r1 = fun(q), l2 = fun(r), r2 = fun(s);
    if (l1 > r1) swap(l1, r1);
    if (l2 > r2) swap(l2, r2);
    x = query(1, 1, n, l1, r1, l2, r2);
    printf("%d\n", x);
    x++;
  }
  return 0;
}