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 = 1000000 + 9;
struct node {
  node *l, *r;
  int s;
  node(int _s = 0) : s(_s) {}
} * root[MAXN];
node *build(int l, int r) {
  node *u = new node;
  if (l < r) {
    int m = (l + r) / 2;
    u->l = build(l, m);
    u->r = build(m + 1, r);
  }
  return u;
}
int N;
void ins(int now, int pos) {
  node *v = root[now - 1];
  node *u = root[now] = new node(now);
  for (int l = 1, r = N; l < r;) {
    int m = (l + r) / 2;
    if (pos > m) {
      u->l = v->l;
      u->r = new node(v->r->s + 1);
      u = u->r, v = v->r;
      l = m + 1;
    } else {
      u->r = v->r;
      u->l = new node(v->l->s + 1);
      u = u->l, v = v->l;
      r = m;
    }
  }
}
int query(node *u, int l, int r, int ql, int qr) {
  if (qr < l || r < ql) return 0;
  if (ql <= l && r <= qr) return u->s;
  int m = (l + r) / 2;
  return query(u->l, l, m, ql, qr) + query(u->r, m + 1, r, ql, qr);
}
int pos[MAXN];
int main() {
  int i, t;
  scanf("%d", &N);
  for (i = 1; i <= N; ++i) {
    scanf("%d", &t);
    pos[t] = i;
  }
  root[0] = build(1, N);
  for (i = 1; i <= N; ++i) {
    scanf("%d", &t);
    ins(i, pos[t]);
  }
  int M, x = 0;
  for (scanf("%d", &M); M--;) {
    int a, b, c, d;
    scanf("%d%d%d%d", &a, &b, &c, &d);
    a = (a - 1 + x) % N + 1, b = (b - 1 + x) % N + 1;
    if (a > b) swap(a, b);
    c = (c - 1 + x) % N + 1, d = (d - 1 + x) % N + 1;
    if (c > d) swap(c, d);
    x = query(root[b], 1, N, c, d) - query(root[a - 1], 1, N, c, d);
    printf("%d\n", x++);
  }
  return 0;
}
