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;
int arr[1 << 20];
int it[22][1 << 20];
int pos[1 << 20];
int LOW, UP;
void makeTree(int l, int r, int lvl) {
  if (l == r) {
    it[lvl][l] = arr[l];
    return;
  }
  int mid = (l + r) / 2;
  makeTree(l, mid, lvl + 1);
  makeTree(mid + 1, r, lvl + 1);
  merge(it[lvl + 1] + l, it[lvl + 1] + mid + 1, it[lvl + 1] + mid + 1,
        it[lvl + 1] + r + 1, it[lvl] + l);
}
int query(int l, int r, int lvl, int f, int t) {
  if (l > t || r < f) return 0;
  if (l >= f && r <= t) {
    int *lw = lower_bound(it[lvl] + l, it[lvl] + r + 1, LOW);
    int *up = upper_bound(it[lvl] + l, it[lvl] + r + 1, UP);
    return up - lw;
  }
  int mid = (l + r) / 2;
  return query(l, mid, lvl + 1, f, t) + query(mid + 1, r, lvl + 1, f, t);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, x;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> x;
    pos[x] = i;
  }
  for (int i = 1; i <= n; i++) {
    cin >> arr[i];
    arr[i] = pos[arr[i]];
  }
  makeTree(1, n, 0);
  x = 0;
  int q;
  cin >> q;
  int a, b, c, d;
  for (int i = 0; i < q; i++) {
    cin >> a >> b >> c >> d;
    a = (a - 1 + x) % n + 1;
    b = (b - 1 + x) % n + 1;
    if (a > b) swap(a, b);
    c = (c - 1 + x) % n + 1;
    d = (d - 1 + x) % n + 1;
    if (c > d) swap(c, d);
    LOW = a;
    UP = b;
    x = query(1, n, 0, c, d);
    cout << x << endl;
    x++;
  }
  return 0;
}