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 MOD = 1000000007;
const int INF = 1e9;
const long long INFLL = 1e18;
const int MAX_N = 100000;
int N, M;
struct Upd {
  int x;
  int y1, y2;
  int type;
  bool operator<(const Upd &a) const { return x < a.x; }
};
vector<Upd> update;
struct Query {
  int x, y;
  int idx;
  bool operator<(const Query &a) const { return x < a.x; }
};
vector<Query> query;
struct S {
  int x, y, z;
};
S arr[MAX_N + 1];
int ans[MAX_N + 1];
struct SEG {
  struct NODE {
    int l, r;
    int sum;
  };
  vector<NODE> seg;
  long long SZ;
  void add() { seg.push_back({-1, -1, 0}); }
  void Init(int x) {
    add();
    SZ = (long long)x;
  }
  void Update(int x, int y) { update(0, 1LL, SZ, x, y); }
  void update(int idx, long long s, long long e, int x, int y) {
    seg[idx].sum += y;
    if (s == e) return;
    if ((long long)x <= (s + e) / 2) {
      if (seg[idx].l == -1) {
        seg[idx].l = seg.size();
        add();
      }
      update(seg[idx].l, s, (s + e) / 2, x, y);
    } else {
      if (seg[idx].r == -1) {
        seg[idx].r = seg.size();
        add();
      }
      update(seg[idx].r, (s + e) / 2 + 1, e, x, y);
    }
  }
  int Calc(int x, int y) { return calc(0, 1LL, SZ, x, y); }
  int calc(int idx, long long s, long long e, int x, int y) {
    if (idx == -1) return 0;
    if ((long long)x <= s && e <= (long long)y) {
      return seg[idx].sum;
    }
    if ((long long)x > e || (long long)y < s) return 0;
    return calc(seg[idx].l, s, (s + e) / 2, x, y) +
           calc(seg[idx].r, (s + e) / 2 + 1, e, x, y);
  }
} Seg1, Seg2;
void input() {
  scanf("%d%d", &N, &M);
  for (int i = 1; i <= N; i++) scanf("%d", &arr[i].x);
  for (int i = 1; i <= N; i++) scanf("%d", &arr[i].y);
  for (int i = 1; i <= N; i++) scanf("%d", &arr[i].z);
  for (int i = 1; i <= N; i++) {
    update.push_back(
        {arr[i].x, arr[i].z + arr[i].x, arr[i].z - arr[i].x + INF, 1});
    update.push_back(
        {arr[i].y + 1, arr[i].z + arr[i].x, arr[i].z - arr[i].x + INF, -1});
  }
  for (int i = 0; i < M; i++) {
    int x;
    scanf("%d", &x);
    query.push_back({x, 0, i});
  }
  for (int i = 0; i < M; i++) {
    int x;
    scanf("%d", &x);
    query[i].y = x;
  }
  sort(update.begin(), update.end());
  sort(query.begin(), query.end());
}
int main() {
  input();
  Seg1.Init(INF * 2);
  Seg2.Init(INF * 2);
  int idx = 0;
  int sum = 0;
  for (int i = 0; i < query.size(); i++) {
    while (idx < update.size() && update[idx].x <= query[i].x) {
      Seg1.Update(update[idx].y1, update[idx].type);
      Seg2.Update(update[idx].y2, update[idx].type);
      sum += update[idx].type;
      idx++;
    }
    int x1 = query[i].x + query[i].y, x2 = query[i].y - query[i].x + INF;
    ans[query[i].idx] = sum - Seg1.Calc(x1 + 1, INF * 2) - Seg2.Calc(1, x2 - 1);
  }
  for (int i = 0; i < M; i++) {
    printf("%d ", ans[i]);
  }
  return 0;
}