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;
const int maxn = 1000;
char buffer[maxn], *S, *T;
inline char Get_Char() {
  if (S == T) {
    T = (S = buffer) + fread(buffer, 1, maxn, stdin);
    if (S == T) return EOF;
  }
  return *S++;
}
int read() {
  char c;
  int re = 0, f = 0;
  for (c = Get_Char(); c < '0' || c > '9'; c = Get_Char()) {
    if (c == '-') f = 1;
  }
  while (c >= '0' && c <= '9') re = re * 10 + (c & 15), c = Get_Char();
  if (f) return -re;
  return re;
}
int n, p[1111111], q[1111111], pos[1111111], m, lans, a, b, c, d, l1, r1, l2,
    r2;
struct hjt {
  int tree[22222222], ls[22222222], rs[22222222], rt[1111111], sz;
  void update(int x, int l, int r, int pre, int &nw) {
    nw = ++sz;
    tree[nw] = tree[pre] + 1;
    ls[nw] = ls[pre];
    rs[nw] = rs[pre];
    if (l == r) return;
    int mid = (l + r) >> 1;
    if (x > mid)
      update(x, mid + 1, r, rs[pre], rs[nw]);
    else
      update(x, l, mid, ls[pre], ls[nw]);
  }
  int query(int beg, int en, int l, int r, int pre, int nw) {
    if (l > en || r < beg) return 0;
    if (l >= beg && r <= en) return tree[nw] - tree[pre];
    int mid = (l + r) >> 1;
    return query(beg, en, l, mid, ls[pre], ls[nw]) +
           query(beg, en, mid + 1, r, rs[pre], rs[nw]);
  }
} ht;
int main() {
  n = read();
  for (int i = 1; i <= n; i++) p[i] = read();
  for (int i = 1; i <= n; i++) q[i] = read();
  for (int i = 1; i <= n; i++) pos[q[i]] = i;
  for (int i = 1; i <= n; i++) {
    ht.update(pos[p[i]], 1, n, ht.rt[i - 1], ht.rt[i]);
  }
  m = read();
  for (int i = 1; i <= m; i++) {
    a = read();
    b = read();
    c = read();
    d = read();
    a = (a - 1 + lans) % n + 1;
    b = (b - 1 + lans) % n + 1;
    c = (c - 1 + lans) % n + 1;
    d = (d - 1 + lans) % n + 1;
    l1 = min(a, b);
    r1 = max(a, b);
    l2 = min(c, d);
    r2 = max(c, d);
    lans = ht.query(l2, r2, 1, n, ht.rt[l1 - 1], ht.rt[r1]);
    printf("%d\n", lans);
    lans++;
  }
  return 0;
}