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;
struct ii {
  int x, y;
  ii(int x = 0, int y = 0) : x(x), y(y) {}
  bool operator<(const ii &a) const {
    if (x != a.x) return x < a.x;
    return y < a.y;
  }
  bool operator==(const ii &a) const { return x == a.x && y == a.y; }
  ii friend operator+(ii a, ii b) { return ii(a.x + b.x, a.y + b.y); }
  ii friend operator-(ii a, ii b) { return ii(a.x - b.x, a.y - b.y); }
};
template <class T>
void sc(T &x) {
  x = 0;
  char c;
  int f = 1;
  while (c = getchar(), c < 48)
    if (c == '-') f = -1;
  do x = x * 10 + (c ^ 48);
  while (c = getchar(), c > 47);
  x *= f;
}
template <class T>
void nt(T x) {
  if (!x) return;
  nt(x / 10);
  putchar(x % 10 + '0');
}
template <class T>
void pt(T x) {
  if (x < 0) putchar('-'), x = -x;
  if (!x)
    putchar('0');
  else
    nt(x);
}
template <class T>
void pts(T x) {
  pt(x);
  putchar(' ');
}
template <class T>
void ptn(T x) {
  pt(x);
  putchar('\n');
}
template <class v>
void pp(v x, int y) {
  static char ch[] = {" \n"};
  pt(x);
  putchar(ch[y]);
}
template <class T>
void PP(T *x, int y) {
  for (int i = 0, i_ = (y)-1; i <= i_; ++i) pp(x[i], i == y - 1);
}
template <class T>
void Max(T &x, T y) {
  if (x < y) x = y;
}
template <class T>
void Min(T &x, T y) {
  if (x > y) x = y;
}
const int maxn = 1000005;
const int maxV = 24000005;
int n, m, tot, lc[maxV], rc[maxV], v[maxV], last, P[maxn], T[maxn];
int f(int z) { return (z + last - 1) % n + 1; }
int insert(int x, int l, int r, int p) {
  int k = ++tot;
  v[k] = v[x] + 1;
  lc[k] = lc[x];
  rc[k] = rc[x];
  if (l < r) {
    int mid = (l + r) >> 1;
    if (p <= mid)
      lc[k] = insert(lc[x], l, mid, p);
    else
      rc[k] = insert(rc[x], mid + 1, r, p);
  }
  return k;
}
int query(int x, int y, int l, int r, int L, int R) {
  if (!y) return 0;
  if (L <= l && r <= R) return v[y] - v[x];
  int mid = (l + r) >> 1, ans = 0;
  if (L <= mid) ans += query(lc[x], lc[y], l, mid, L, R);
  if (R > mid) ans += query(rc[x], rc[y], mid + 1, r, L, R);
  return ans;
}
int end;
int main() {
  sc(n);
  int u;
  for (int i = 1, i_ = (n + 1) - 1; i <= i_; ++i) sc(u), T[u] = i;
  for (int i = 1, i_ = (n + 1) - 1; i <= i_; ++i) sc(u), P[T[u]] = i;
  for (int i = 1, i_ = (n + 1) - 1; i <= i_; ++i)
    T[i] = insert(T[i - 1], 1, n, P[i]);
  sc(m);
  while (m--) {
    int l, r, a, b;
    sc(l);
    sc(r);
    sc(a);
    sc(b);
    l = f(l);
    r = f(r);
    a = f(a);
    b = f(b);
    if (l > r) swap(l, r);
    if (a > b) swap(a, b);
    last = query(T[l - 1], T[r], 1, n, a, b);
    ptn(last);
    last++;
  }
  return 0;
}