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;
const int maxn = 1000005;
int n, m, tot = 0, root[maxn], key, p[maxn];
int c[maxn * 30][2], sum[maxn * 30];
int read() {
  char ch = getchar();
  while (ch < '0' || ch > '9') ch = getchar();
  int res = 0;
  while (ch >= '0' && ch <= '9') res = res * 10 + ch - '0', ch = getchar();
  return res;
}
void build(int &x, int L, int R) {
  if (!x) x = ++tot;
  if (L == R) return;
  build(c[x][0], L, ((L + R) >> 1));
  build(c[x][1], ((L + R) >> 1) + 1, R);
}
int update(int x, int L, int R, int index) {
  int now = ++tot;
  if (L == R) return sum[now] = sum[x] + 1, now;
  if (index <= ((L + R) >> 1))
    c[now][0] = update(c[x][0], L, ((L + R) >> 1), index), c[now][1] = c[x][1];
  else
    c[now][1] = update(c[x][1], ((L + R) >> 1) + 1, R, index),
    c[now][0] = c[x][0];
  sum[now] = sum[c[now][0]] + sum[c[now][1]];
}
int query(int x1, int x2, int L, int R, int QL, int QR) {
  if (QL > R || QR < L) return 0;
  if (QL <= L && QR >= R) return sum[x2] - sum[x1];
  return query(c[x1][0], c[x2][0], L, ((L + R) >> 1), QL, QR) +
         query(c[x1][1], c[x2][1], ((L + R) >> 1) + 1, R, QL, QR);
}
int f(int x) { return (x - 1 + key + 1) % n + 1; }
int main() {
  n = read();
  for (int i = 1, x; i <= n; i++) x = read(), p[x] = i;
  build(root[0], 1, n);
  for (int i = 1, x; i <= n; i++)
    x = read(), root[i] = update(root[i - 1], 1, n, p[x]);
  m = read(), key = -1;
  for (int i = 1; i <= m; i++) {
    int a = read(), b = read(), c = read(), d = read();
    int fa = f(a), fb = f(b), fc = f(c), fd = f(d);
    int L1 = min(fa, fb), R1 = max(fa, fb), L2 = min(fc, fd), R2 = max(fc, fd);
    printf("%d\n", key = query(root[L2 - 1], root[R2], 1, n, L1, R1));
  }
  return 0;
}
