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 = 1e6 + 5;
int N, M, p[MAXN], q[MAXN], x[MAXN], lastans = -1;
int rt[MAXN], sum[MAXN << 2], ls[MAXN << 2], rs[MAXN << 2], Tlen;
void ins(int o, int old, int l, int r, int pos, int k) {
  if (l == r)
    sum[o] = sum[old] + k;
  else {
    ls[o] = ls[old], rs[o] = rs[old];
    if (pos <= ((l + r) >> 1))
      ls[o] = ++Tlen, ins(ls[o], ls[old], l, ((l + r) >> 1), pos, k);
    else
      rs[o] = ++Tlen, ins(rs[o], rs[old], ((l + r) >> 1) + 1, r, pos, k);
    sum[o] = sum[ls[o]] + sum[rs[o]];
  }
}
int qry(int o, int l, int r, int L, int R) {
  if (!o) return 0;
  if (l == L && r == R)
    return sum[o];
  else {
    if (R <= ((l + r) >> 1))
      return qry(ls[o], l, ((l + r) >> 1), L, R);
    else if (L > ((l + r) >> 1))
      return qry(rs[o], ((l + r) >> 1) + 1, r, L, R);
    else
      return qry(ls[o], l, ((l + r) >> 1), L, ((l + r) >> 1)) +
             qry(rs[o], ((l + r) >> 1) + 1, r, ((l + r) >> 1) + 1, R);
  }
}
int main() {
  scanf("%d", &N);
  for (int i = 1; i <= N; ++i) scanf("%d", &p[i]);
  for (int i = 1; i <= N; ++i) scanf("%d", &q[i]), x[q[i]] = i;
  for (int i = 1; i <= N; ++i)
    rt[i] = ++Tlen, ins(rt[i], rt[i - 1], 1, N, x[p[i]], 1);
  scanf("%d", &M);
  while (M--) {
    int l1, r1, l2, r2;
    scanf("%d%d%d%d", &l1, &r1, &l2, &r2);
    l1 = (l1 + lastans) % N + 1, r1 = (r1 + lastans) % N + 1,
    l2 = (l2 + lastans) % N + 1, r2 = (r2 + lastans) % N + 1;
    if (l1 > r1) swap(l1, r1);
    if (l2 > r2) swap(l2, r2);
    lastans = qry(rt[r1], 1, N, l2, r2) - qry(rt[l1 - 1], 1, N, l1, r1);
    printf("%d\n", lastans);
  }
  return 0;
}
