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>
char In[1 << 22], *p1 = In, *p2 = In, Ch, Out[1 << 21], *Outp = Out, St[20],
                  *Tp = St;
inline int Getint(int x = 0) {
  while (!isdigit(
      Ch =
          (p1 == p2 && (p2 = (p1 = In) + fread(In, 1, 1 << 22, stdin), p1 == p2)
               ? EOF
               : *p1++)))
    ;
  for (; isdigit(Ch);
       Ch = (p1 == p2 &&
                     (p2 = (p1 = In) + fread(In, 1, 1 << 22, stdin), p1 == p2)
                 ? EOF
                 : *p1++))
    x = x * 10 + (Ch ^ 48);
  return x;
}
inline void Putint(int x) {
  do *Tp++ = x % 10 ^ 48;
  while (x /= 10);
  do *Outp++ = *--Tp;
  while (Tp != St);
  *Outp++ = '\n';
}
const int N = 1000005, M = 22;
int n, m, A[N], B[N];
int Root[N], Sum[N * M], Lc[N * M], Rc[N * M], Tn;
int o, L, R;
void Insert(int &p, int ps, int l, int r) {
  if (p = ++Tn, l == r) {
    Sum[p] = Sum[ps] + 1;
    return;
  }
  const int Mid = (l + r) >> 1;
  if (o <= Mid)
    Insert(Lc[p], Lc[ps], l, Mid), Rc[p] = Rc[ps];
  else
    Insert(Rc[p], Rc[ps], Mid + 1, r), Lc[p] = Lc[ps];
  Sum[p] = Sum[Lc[p]] + Sum[Rc[p]];
}
int Query(int p, int l, int r) {
  if (L <= l && r <= R) return Sum[p];
  int Mid = (l + r) >> 1, Res = 0;
  if (L <= Mid) Res = Query(Lc[p], l, Mid);
  if (R > Mid) Res += Query(Rc[p], Mid + 1, r);
  return Res;
}
int main() {
  n = Getint();
  for (register int i = 1; i <= n; ++i) A[Getint()] = i;
  for (register int i = 1; i <= n; ++i)
    o = B[i] = A[Getint()], Insert(Root[i], Root[i - 1], 1, n);
  m = Getint();
  for (register int a, b, c, d, x = -1; m--;) {
    a = (((Getint()) + x) % n + 1), b = (((Getint()) + x) % n + 1),
    c = (((Getint()) + x) % n + 1), d = (((Getint()) + x) % n + 1);
    if (a > b) std::swap(a, b);
    if (c > d) std::swap(c, d);
    L = a, R = b, Putint(x = Query(Root[d], 1, n) - Query(Root[c - 1], 1, n));
  }
  return fwrite(Out, 1, Outp - Out, stdout), 0;
}