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;
template <typename T>
T read1() {
  T t = 0;
  char k;
  bool vis = 0;
  do (k = getchar()) == '-' && (vis = 1);
  while ('0' > k || k > '9');
  while ('0' <= k && k <= '9')
    t = (t << 3) + (t << 1) + (k ^ '0'), k = getchar();
  return vis ? -t : t;
}
struct node {
  node *l, *r;
  int v;
  node() {
    l = r = NULL;
    v = 0;
  }
} * root[1000005];
int s, m, a[1000005];
node *insert(node *n, int x, int l = 1, int r = 1000000) {
  node *v = new node;
  if (n) v->l = n->l, v->r = n->r, v->v = n->v;
  ++v->v;
  if (l == r) return v;
  int mid = l + r >> 1;
  if (x <= mid)
    v->l = insert(v->l, x, l, mid);
  else
    v->r = insert(v->r, x, mid + 1, r);
  return v;
}
int value(node *x) { return x ? x->v : 0; }
node *L(node *x) { return x ? x->l : NULL; }
node *R(node *x) { return x ? x->r : NULL; }
int query(node *x, node *y, int lx, int rx, int l = 1, int r = 1000000) {
  if (l == lx && r == rx) return value(y) - value(x);
  if (!x && !y) return 0;
  int mid = l + r >> 1;
  if (rx <= mid) return query(L(x), L(y), lx, rx, l, mid);
  if (lx > mid) return query(R(x), R(y), lx, rx, mid + 1, r);
  return query(L(x), L(y), lx, mid, l, mid) +
         query(R(x), R(y), mid + 1, rx, mid + 1, r);
}
int main() {
  s = read1<long long>();
  for (int i = 1; i <= s; ++i) a[read1<long long>()] = i;
  for (int i = 1; i <= s; ++i)
    root[i] = insert(root[i - 1], a[read1<long long>()]);
  int la = 0;
  for (int T = read1<long long>(); T--;) {
    int l1 = (read1<long long>() + la - 1) % s + 1,
        r1 = (read1<long long>() + la - 1) % s + 1,
        l2 = (read1<long long>() + la - 1) % s + 1,
        r2 = (read1<long long>() + la - 1) % s + 1;
    if (l1 > r1) swap(l1, r1);
    if (l2 > r2) swap(l2, r2);
    printf("%d\n", la = query(root[l2 - 1], root[r2], l1, r1));
    ++la;
  }
  return 0;
}