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;
inline bool SR(int &x) { return scanf("%d", &x) == 1; }
inline bool SR(long long &x) { return scanf("%lld", &x) == 1; }
inline bool SR(double &x) { return scanf("%lf", &x); }
inline bool SR(char *s) { return scanf("%s", s); }
inline bool RI() { return 1; }
template <typename I, typename... T>
bool RI(I &x, T &...tail) {
  return SR(x) && RI(tail...);
}
inline void SP(const int x) { printf("%d", x); }
inline void SP(const long long x) { printf("%lld", x); }
inline void SP(const double x) { printf("%.10f", x); }
inline void SP(const char *s) { printf("%s", s); }
inline void PL() { puts(""); }
template <typename I, typename... T>
void PL(const I x, const T... tail) {
  SP(x);
  if (sizeof...(tail)) putchar(' ');
  PL(tail...);
}
template <typename I>
inline void _DOING(const char *s, I &&x) {
  cerr << s << " = " << x << '\n';
}
template <typename I, typename... T>
void _DOING(const char *s, I &&x, T &&...tail) {
  int c = 0;
  while (*s != ',' || c != 0) {
    if (*s == '(' || *s == '[' || *s == '{') c++;
    if (*s == ')' || *s == ']' || *s == '}') c--;
    cerr << *s++;
  }
  cerr << " = " << x << " , ";
  _DOING(s + 1, tail...);
}
const int INF = 0x3f3f3f3f;
const long long INF64 = 0x3f3f3f3f3f3f3f3fll;
inline int RAND() {
  static int x = 880301;
  return x = (x * 0xdefacedll + 1ll) % 0x7fffffffll;
}
const int MAX_N = 1000000 + 7;
struct Segt {
  Segt *lc, *rc;
  int l, r, sz;
  Segt(int _l = 0, int _r = 0) : lc(0), rc(0), l(_l), r(_r), sz(0) {}
};
Segt *build(int l, int r) {
  Segt *o = new Segt(l, r);
  if (l != r) {
    int m = (l + r) >> 1;
    o->lc = build(l, m);
    o->rc = build(m + 1, r);
  }
  return o;
}
Segt *insert(Segt *o, int p) {
  o = new Segt(*o);
  int l = o->l;
  int r = o->r;
  o->sz++;
  if (l == r) return o;
  int m = (l + r) >> 1;
  if (p <= m)
    o->lc = insert(o->lc, p);
  else
    o->rc = insert(o->rc, p);
  return o;
}
int query(Segt *a, Segt *b, int ql, int qr) {
  int l = a->l;
  int r = a->r;
  if (ql <= l && r <= qr) return b->sz - a->sz;
  int m = (l + r) >> 1;
  int res = 0;
  if (ql <= m) res += query(a->lc, b->lc, ql, qr);
  if (m < qr) res += query(a->rc, b->rc, ql, qr);
  return res;
}
Segt *tr[MAX_N];
int P[MAX_N], Q[MAX_N], A[MAX_N];
int main() {
  int n;
  RI(n);
  for (int i = (1); i <= int(n); i++) {
    int x;
    RI(x);
    P[x] = i;
  }
  for (int i = (1); i <= int(n); i++) {
    int x;
    RI(x);
    Q[x] = i;
  }
  for (int i = (1); i <= int(n); i++) A[P[i]] = Q[i];
  tr[0] = build(1, n);
  for (int i = (1); i <= int(n); i++) tr[i] = insert(tr[i - 1], A[i]);
  int m;
  RI(m);
  int ans = 0;
  while (m--) {
    int a, b, c, d;
    RI(a, b, c, d);
    a = ((a - 1 + ans) % n) + 1;
    b = ((b - 1 + ans) % n) + 1;
    c = ((c - 1 + ans) % n) + 1;
    d = ((d - 1 + ans) % n) + 1;
    int l1 = min(a, b);
    int r1 = max(a, b);
    int l2 = min(c, d);
    int r2 = max(c, d);
    ans = query(tr[l1 - 1], tr[r1], l2, r2);
    PL(ans);
  }
  return 0;
}
