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
1
1
1
1
1 1 1 1

1

#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 S = 1;
int n, m;
int md[maxn], end, last, A[maxn], B[maxn], T[maxn], P[maxn];
int Ans[1005][1005];
int f(int z) { return (z + last - 1) % n + 1; }
void get(int &l, int &r) {
  int long long = l / S;
  if (l != long long *S) long long ++;
  int rr = r / S;
  if (r != rr * S + S - 1) rr--;
  l = long long;
  r = rr;
}
int solve(int l, int r, int low, int high) {
  int ans = 0;
  if (l / S == r / S) {
    for (int i = l, i_ = (r + 1) - 1; i <= i_; ++i)
      ans += low <= P[i] && P[i] <= high;
    return ans;
  }
  int long long = l, rr = r;
  get(long long, rr);
  int xx = low, yy = high;
  get(xx, yy);
  if (xx <= yy)
    for (int i = long long, i_ = (rr + 1) - 1; i <= i_; ++i) {
      ans += Ans[i][yy];
      if (xx) ans -= Ans[i][xx - 1];
    }
  for (int i = l, i_ = (long long *S) - 1; i <= i_; ++i)
    ans += low <= P[i] && P[i] <= high;
  for (int i = rr * S + S, i_ = (r + 1) - 1; i <= i_; ++i)
    ans += low <= P[i] && P[i] <= high;
  return ans;
}
int main() {
  freopen("chk2.out", "w", stdout);
  sc(n);
  for (int i = 0, i_ = (n)-1; i <= i_; ++i) sc(A[i]), T[A[i]] = i;
  for (int i = 0, i_ = (n)-1; i <= i_; ++i) sc(B[i]), P[T[B[i]]] = i;
  for (int cas = 0, cas_ = (n / S + 1) - 1; cas <= cas_; ++cas) {
    int l = cas * S, r = min(n, l + S);
    for (int i = l, i_ = (r)-1; i <= i_; ++i) md[A[i]] = 1;
    for (int i = 0, i_ = (n / S + 1) - 1; i <= i_; ++i) {
      int long long = i * S, rr = min(n, long long + S);
      for (int j = long long, j_ = (rr)-1; j <= j_; ++j)
        Ans[cas][i] += md[B[j]];
      if (i) Ans[cas][i] += Ans[cas][i - 1];
    }
    for (int i = l, i_ = (r)-1; i <= i_; ++i) md[A[i]] = 0;
  }
  sc(m);
  while (m--) {
    int l, r, low, high;
    sc(l);
    l = f(l);
    sc(r);
    r = f(r);
    if (l > r) swap(l, r);
    sc(low);
    low = f(low);
    sc(high);
    high = f(high);
    if (low > high) swap(low, high);
    last = solve(l - 1, r - 1, low - 1, high - 1);
    ptn(last);
    last++;
  }
  return 0;
}
