Problem: There are m people living in a city. There are n dishes sold in the city. Each dish i has a price p_i, a standard s_i and a beauty b_i. Each person j has an income of inc_j and a preferred beauty pref_j. 

A person would never buy a dish whose standard is less than the person's income. Also, a person can't afford a dish with a price greater than the income of the person. In other words, a person j can buy a dish i only if p_i ≤ inc_j ≤ s_i.

Also, a person j can buy a dish i, only if |b_i-pref_j| ≤ (inc_j-p_i). In other words, if the price of the dish is less than the person's income by k, the person will only allow the absolute difference of at most k between the beauty of the dish and his/her preferred beauty. 

Print the number of dishes that can be bought by each person in the city.

Input

The first line contains two integers n and m (1 ≤ n ≤ 10^5, 1 ≤ m ≤ 10^5), the number of dishes available in the city and the number of people living in the city.

The second line contains n integers p_i (1 ≤ p_i ≤ 10^9), the price of each dish.

The third line contains n integers s_i (1 ≤ s_i ≤ 10^9), the standard of each dish.

The fourth line contains n integers b_i (1 ≤ b_i ≤ 10^9), the beauty of each dish.

The fifth line contains m integers inc_j (1 ≤ inc_j ≤ 10^9), the income of every person.

The sixth line contains m integers pref_j (1 ≤ pref_j ≤ 10^9), the preferred beauty of every person.

It is guaranteed that for all integers i from 1 to n, the following condition holds: p_i ≤ s_i.

Output

Print m integers, the number of dishes that can be bought by every person living in the city.

Examples

Input


3 3
2 1 3
2 4 4
2 1 1
2 2 3
1 2 4


Output


1 2 0 

Input


4 3
1 2 1 1
3 3 1 3
2 1 3 2
1 1 3
1 2 1


Output


0 2 3 

Note

In the first example, the first person can buy dish 2, the second person can buy dishes 1 and 2 and the third person can buy no dishes.

In the second example, the first person can buy no dishes, the second person can buy dishes 1 and 4, and the third person can buy dishes 1, 2 and 4.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int maxn = 100005;
int n, m, p[maxn], s[maxn], b[maxn], inc[maxn], pref[maxn], ans[maxn], tot;
pair<int, int> t[maxn * 3];
int Id[maxn], rk[maxn];
inline int gi() {
  char c = getchar();
  while (c < '0' || c > '9') c = getchar();
  int sum = 0;
  while ('0' <= c && c <= '9') sum = sum * 10 + c - 48, c = getchar();
  return sum;
}
int rt[maxn * 6];
namespace sgt {
int lch[maxn * 200], rch[maxn * 200], sum[maxn * 200], tot;
int Id[maxn], rk[maxn];
inline bool cmp(int x, int y) { return p[x] - b[x] < p[y] - b[y]; }
void pre() {
  for (int i = 1; i <= n; ++i) Id[i] = i;
  sort(Id + 1, Id + n + 1, cmp);
  for (int i = 1; i <= n; ++i) rk[Id[i]] = i, Id[i] = p[Id[i]] - b[Id[i]];
}
void insert(int &s, int l, int r, int x, int v) {
  if (!s) s = ++tot;
  sum[s] += v;
  if (l == r) return;
  if (x <= ((l + r) >> 1))
    insert(lch[s], l, ((l + r) >> 1), x, v);
  else
    insert(rch[s], ((l + r) >> 1) + 1, r, x, v);
}
int query(int &s, int l, int r, int x) {
  if (!s || r <= x) return sum[s];
  if (x <= ((l + r) >> 1))
    return query(lch[s], l, ((l + r) >> 1), x);
  else
    return query(lch[s], l, ((l + r) >> 1), x) +
           query(rch[s], ((l + r) >> 1) + 1, r, x);
}
int query(int s, int x) {
  int pos = upper_bound(Id + 1, Id + n + 1, x) - Id - 1;
  return query(rt[s], 1, n, pos);
}
};  // namespace sgt
void insert(int s, int l, int r, int x, int y, int tp) {
  sgt::insert(rt[s], 1, n, sgt::rk[y], tp);
  if (l == r) return;
  if (x <= ((l + r) >> 1))
    insert((s << 1), l, ((l + r) >> 1), x, y, tp);
  else
    insert((s << 1 | 1), ((l + r) >> 1) + 1, r, x, y, tp);
}
int query(int s, int l, int r, int x, int y) {
  if (r <= x) return sgt::query(s, y);
  if (x <= ((l + r) >> 1))
    return query((s << 1), l, ((l + r) >> 1), x, y);
  else
    return query((s << 1), l, ((l + r) >> 1), x, y) +
           query((s << 1 | 1), ((l + r) >> 1) + 1, r, x, y);
}
void insert(int x, int tp) { insert(1, 1, n, rk[x], x, tp); }
int query(int x) {
  int pos = upper_bound(Id + 1, Id + n + 1, inc[x] + pref[x]) - Id - 1;
  return pos ? query(1, 1, n, pos, inc[x] - pref[x]) : 0;
}
inline bool cmp(int x, int y) { return p[x] + b[x] < p[y] + b[y]; }
int main() {
  n = gi();
  m = gi();
  for (int i = 1; i <= n; ++i) p[i] = gi();
  for (int i = 1; i <= n; ++i) s[i] = gi();
  for (int i = 1; i <= n; ++i) b[i] = gi();
  for (int i = 1; i <= m; ++i) inc[i] = gi();
  for (int i = 1; i <= m; ++i) pref[i] = gi();
  for (int i = 1; i <= n; ++i) Id[i] = i;
  sort(Id + 1, Id + n + 1, cmp);
  for (int i = 1; i <= n; ++i) rk[Id[i]] = i, Id[i] = p[Id[i]] + b[Id[i]];
  sgt::pre();
  for (int i = 1; i <= n; ++i) t[++tot] = make_pair(p[i], i);
  for (int i = 1; i <= m; ++i) t[++tot] = make_pair(inc[i], i + n);
  for (int i = 1; i <= n; ++i) t[++tot] = make_pair(s[i], i + n + m);
  sort(t + 1, t + tot + 1);
  for (int i = 1; i <= tot; ++i) {
    if (t[i].second <= n)
      insert(t[i].second, 1);
    else if (t[i].second <= n + m)
      ans[t[i].second - n] = query(t[i].second - n);
    else
      insert(t[i].second - n - m, -1);
  }
  for (int i = 1; i <= m; ++i) printf("%d ", ans[i]);
  return 0;
}