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;
struct Dish {
  Dish(int p = 0, int s = 0, int b = 0) : price(p), standard(s), beauty(b) {}
  int price;
  int beauty;
  int standard;
  bool operator<(const Dish& other) const { return beauty < other.beauty; }
};
struct Person {
  Person(int i = 0, int p = 0, int id = 0) : income(i), preferred(p), idx(id) {}
  int idx;
  int income;
  int preferred;
  bool operator<(const Person& person) const {
    return preferred < person.preferred;
  }
};
class FenwickTree {
 public:
  FenwickTree(int n) : N(n), data(N, 0) {}
  void reset() { fill(data.begin(), data.end(), 0); }
  void add(int x, int v) {
    for (; x < N; x |= x + 1) {
      data[x] += v;
    }
  }
  int get(int x) {
    int res = 0;
    for (; x >= 0; x = (x & (x + 1)) - 1) {
      res += data[x];
    }
    return res;
  }

 private:
  int N;
  vector<int> data;
};
class Solution {
 public:
  vector<int> calc(vector<Dish>& dishes, vector<Person>& persons) {
    int n = dishes.size();
    int m = persons.size();
    vector<int> res(m, 0);
    vector<pair<int, int>> events;
    int idx = 1;
    vector<int> sumBP;
    vector<int> diffBP;
    for (auto& dish : dishes) {
      events.emplace_back(dish.price, idx);
      events.emplace_back(dish.standard + 1, -idx);
      ++idx;
      sumBP.emplace_back(dish.price + dish.beauty);
      diffBP.emplace_back(dish.price - dish.beauty);
    }
    int N = n + 1;
    idx = 0;
    for (auto& person : persons) {
      events.emplace_back(person.income, N + idx);
      ++idx;
    }
    sort(events.begin(), events.end());
    auto shrink = [&](vector<int>& nums) {
      sort(nums.begin(), nums.end());
      nums.resize(unique(nums.begin(), nums.end()) - nums.begin());
    };
    shrink(sumBP);
    shrink(diffBP);
    auto lindex = [&](vector<int>& nums, int num) -> int {
      int res = lower_bound(nums.begin(), nums.end(), num) - nums.begin();
      return res;
    };
    auto uindex = [&](vector<int>& nums, int num) -> int {
      int res = upper_bound(nums.begin(), nums.end(), num) - nums.begin();
      --res;
      return res;
    };
    FenwickTree sft(sumBP.size() + 100);
    FenwickTree dft(diffBP.size() + 100);
    int cnt = 0;
    for (auto& event : events) {
      int p = event.first;
      int id = event.second;
      if (id < 0) {
        int i = -id - 1;
        id = lindex(sumBP, dishes[i].price + dishes[i].beauty);
        sft.add(id, -1);
        id = lindex(diffBP, dishes[i].price - dishes[i].beauty);
        dft.add(id, -1);
        --cnt;
      } else if (id < N) {
        --id;
        int i = id;
        id = lindex(sumBP, dishes[i].price + dishes[i].beauty);
        sft.add(id, 1);
        id = lindex(diffBP, dishes[i].price - dishes[i].beauty);
        dft.add(id, 1);
        ++cnt;
      } else {
        int i = id - N;
        id = uindex(sumBP, persons[i].income + persons[i].preferred);
        int ans = sft.get(id);
        id = uindex(diffBP, persons[i].income - persons[i].preferred);
        ans += dft.get(id);
        ans -= cnt;
        res[i] += ans;
      }
    }
    return res;
  }
};
int main(int argc, char** argv) {
  ios::sync_with_stdio(false);
  cin.tie(0);
  Solution sol;
  int n, m;
  cin >> n >> m;
  vector<Dish> dishes(n);
  for (int i = 0; i < n; ++i) {
    cin >> dishes[i].price;
  }
  for (int i = 0; i < n; ++i) {
    cin >> dishes[i].standard;
  }
  for (int i = 0; i < n; ++i) {
    cin >> dishes[i].beauty;
  }
  vector<Person> persons(m);
  for (int i = 0; i < m; ++i) {
    cin >> persons[i].income;
  }
  for (int i = 0; i < m; ++i) {
    cin >> persons[i].preferred;
  }
  for (int i = 0; i < m; ++i) {
    persons[i].idx = i;
  }
  auto res = sol.calc(dishes, persons);
  for (auto r : res) {
    cout << r << " ";
  }
  cout << "\n";
  return 0;
}