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>
#pragma GCC optimize(3)
#pragma warning(disable : 4996)
template <typename T>
T min(T x, T y) {
  return x < y ? x : y;
}
template <typename T>
T max(T x, T y) {
  return x > y ? x : y;
};
const int MAXN = 100005;
struct event {
  int ty;
  int id;
  int add;
  int w;
  event(int T = 0, int I = 0, int A = 0, int W = 0) {
    ty = T;
    id = I;
    add = A;
    w = W;
  }
  bool operator<(event t) const {
    if (w != t.w) return w < t.w;
    return ty < t.ty;
  }
};
int N, M;
int p[MAXN], s[MAXN], b[MAXN];
int inc[MAXN], pre[MAXN];
event e[MAXN * 5];
int cnt, num, rank[MAXN * 5], tmp[MAXN * 5];
int LT[MAXN * 5], RT[MAXN * 5];
int ans[MAXN];
int find(int x) {
  int pos = std::lower_bound(tmp, tmp + num, x) - tmp;
  return rank[pos];
}
void modify(int *M, int id, int k) {
  for (int i = id; i < 5 * MAXN; i += ((i) & -(i))) M[i] += k;
}
int query(int *M, int id) {
  int ans = 0;
  for (int i = id; i; i -= ((i) & -(i))) ans += M[i];
  return ans;
}
void solve() {
  for (int i = 1; i <= cnt; i++) {
    if (e[i].ty == 1)
      modify(LT, find(e[i].id), e[i].add);
    else if (e[i].ty == 2)
      modify(RT, find(e[i].id), e[i].add);
    else {
      int id = e[i].id;
      ans[id] += query(LT, find(inc[id] + pre[id]));
      ans[id] += query(RT, find(-inc[id] + pre[id]));
    }
  }
  for (int i = 1; i <= M; i++) printf("%d ", ans[i]);
}
void init() {
  scanf("%d %d", &N, &M);
  for (int i = 1; i <= N; i++) scanf("%d", p + i);
  for (int i = 1; i <= N; i++) scanf("%d", s + i);
  for (int i = 1; i <= N; i++) scanf("%d", b + i);
  for (int i = 1; i <= M; i++) scanf("%d", inc + i);
  for (int i = 1; i <= M; i++) scanf("%d", pre + i);
  for (int i = 1; i <= N; i++) {
    e[++cnt] = event(1, p[i] + b[i], 1, p[i]);
    e[++cnt] = event(1, p[i] + b[i], -1, s[i] + 1);
    e[++cnt] = event(2, -p[i] + 1 + b[i], -1, p[i]);
    e[++cnt] = event(2, -p[i] + 1 + b[i], 1, s[i] + 1);
    tmp[num++] = p[i] + b[i];
    tmp[num++] = -p[i] + 1 + b[i];
  }
  for (int i = 1; i <= M; i++) {
    e[++cnt] = event(3, i, 0, inc[i]);
    tmp[num++] = inc[i] + pre[i];
    tmp[num++] = -inc[i] + pre[i];
  }
  std::sort(e + 1, e + cnt + 1);
  std::sort(tmp, tmp + num);
  int cur = 0;
  for (int i = 0; i < num; i++) {
    if (!i || tmp[i] > tmp[i - 1]) cur++;
    rank[i] = cur;
  }
}
int main() {
  init();
  solve();
  return 0;
}