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 N = 1111111;
int mx[N << 2], mn[N << 2], rec[N << 2];
int nid[N];
void up(int rt) {
  mx[rt] = max(mx[rt << 1], mx[rt << 1 | 1]);
  mn[rt] = min(mn[rt << 1], mn[rt << 1 | 1]);
}
void input(int n) {
  int x;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &x);
    nid[x] = i;
  }
}
void build(int l, int r, int rt) {
  if (l == r) {
    int x;
    scanf("%d", &x);
    mx[rt] = mn[rt] = nid[x];
    rec[rt] = x;
    return;
  }
  int m = l + r >> 1;
  build(l, m, rt << 1);
  build(m + 1, r, rt << 1 | 1);
  up(rt);
}
bool cross(int a, int b, int c, int d) {
  if (d < a) return false;
  if (b < c) return false;
  return true;
}
int query(int L, int R, int A, int B, int l, int r, int rt) {
  if (l >= r) return rec[rt];
  int m = l + r >> 1;
  if (A <= l && r <= B) {
    if (cross(mn[rt << 1], mx[rt << 1], L, R))
      return query(L, R, A, B, l, m, rt << 1);
    if (cross(mn[rt << 1 | 1], mx[rt << 1 | 1], L, R))
      return query(L, R, A, B, m + 1, r, rt << 1 | 1);
    return -1;
  } else {
    int ans = -1;
    if (A <= m) ans = query(L, R, A, B, l, m, rt << 1);
    if (ans == -1 && m < B) ans = query(L, R, A, B, m + 1, r, rt << 1 | 1);
    return ans;
  }
}
int main() {
  int n;
  while (cin >> n) {
    input(n);
    build(1, n, 1);
    int ans = 0;
    int m, a, b, c, d;
    cin >> m;
    while (m--) {
      scanf("%d%d%d%d", &a, &b, &c, &d);
      int na = (a + ans - 1) % n + 1;
      int nb = (b + ans - 1) % n + 1;
      int nc = (c + ans - 1) % n + 1;
      int nd = (d + ans - 1) % n + 1;
      ans = query(min(na, nb), max(na, nb), min(nc, nd), max(nc, nd), 1, n, 1);
      printf("%d\n", ans);
    }
  }
  return 0;
}
