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 + 7;
vector<int> seg[N * 4];
int n, m;
int pinf[N];
int a[N];
void fil(int p, int val, int v = 1, int s = 1, int e = n + 1) {
  seg[v].push_back(val);
  if (e - s == 1) return;
  if (p < ((s + e) / 2))
    fil(p, val, v * 2, s, ((s + e) / 2));
  else
    fil(p, val, v * 2 + 1, ((s + e) / 2), e);
}
int get(int l, int r, int lw, int hg, int v = 1, int s = 1, int e = n + 1) {
  if (r <= s || e <= l) return 0;
  if (l <= s && e <= r)
    return upper_bound(seg[v].begin(), seg[v].end(), hg) -
           lower_bound(seg[v].begin(), seg[v].end(), lw);
  return get(l, r, lw, hg, v * 2, s, ((s + e) / 2)) +
         get(l, r, lw, hg, v * 2 + 1, ((s + e) / 2), e);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int t;
    scanf("%d", &t);
    pinf[t] = i;
  }
  for (int i = 1; i <= n; i++) {
    int t;
    scanf("%d", &t);
    a[i] = pinf[t];
    fil(i, a[i]);
  }
  for (int i = 1; i < 4 * N; i++) sort(seg[i].begin(), seg[i].end());
  scanf("%d", &m);
  int lstans = -1;
  while (m--) {
    lstans++;
    int l, r, s, e;
    scanf("%d%d%d%d", &l, &r, &s, &e);
    l = ((l - 1 + lstans) % n) + 1;
    r = ((r - 1 + lstans) % n) + 1;
    s = ((s - 1 + lstans) % n) + 1;
    e = ((e - 1 + lstans) % n) + 1;
    if (r < l) swap(l, r);
    if (e < s) swap(s, e);
    lstans = get(s, e + 1, l, r);
    printf("%d\n", lstans);
  }
  return 0;
}