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>
struct Event {
  int pos, left, right, type, id;
};
const int MAXN = 1e5;
const int MAXPOP = 3e5;
const int MAXIND = 2e5 + 1;
int price[MAXN], standard[MAXN], beauty[MAXN];
int income[MAXN], preferred[MAXN], canBuy[MAXN];
Event city[MAXPOP];
int leftTree[MAXIND];
int rightTree[MAXIND];
void modif(int *tree, int index, int val) {
  while (index <= MAXIND) {
    tree[index] += val;
    index += index & (-index);
  }
}
int sum(int *tree, int index) {
  int result = 0;
  while (index > 0) {
    result += tree[index];
    index -= index & (-index);
  }
  return result;
}
int main() {
  std::ios::sync_with_stdio(false);
  std::cout.tie(nullptr);
  std::cin.tie(nullptr);
  int dishes, people;
  std::cin >> dishes >> people;
  int population = 2 * dishes + people;
  for (int iDish = 0; iDish < dishes; iDish++) std::cin >> price[iDish];
  for (int iDish = 0; iDish < dishes; iDish++) std::cin >> standard[iDish];
  for (int iDish = 0; iDish < dishes; iDish++) std::cin >> beauty[iDish];
  for (int iPers = 0; iPers < people; iPers++) std::cin >> income[iPers];
  for (int iPers = 0; iPers < people; iPers++) std::cin >> preferred[iPers];
  for (int iDish = 0; iDish < dishes; iDish++) {
    city[2 * iDish].pos = price[iDish];
    city[2 * iDish + 1].pos = standard[iDish];
    city[2 * iDish].left = city[2 * iDish + 1].left =
        beauty[iDish] - price[iDish];
    city[2 * iDish].right = city[2 * iDish + 1].right =
        beauty[iDish] + price[iDish];
    city[2 * iDish].type = 1;
    city[2 * iDish + 1].type = -1;
    city[2 * iDish].id = city[2 * iDish + 1].id = -1;
  }
  for (int iPers = 0; iPers < people; iPers++) {
    city[iPers + 2 * dishes].pos = income[iPers];
    city[iPers + 2 * dishes].left = preferred[iPers] - income[iPers] - 1;
    city[iPers + 2 * dishes].right = preferred[iPers] + income[iPers] + 1;
    city[iPers + 2 * dishes].type = 0;
    city[iPers + 2 * dishes].id = iPers;
  }
  int cur = 0;
  std::sort(city, city + population,
            [](const Event &a, const Event &b) { return (a.left < b.left); });
  for (int iLeft = 1; cur < population; iLeft++) {
    int val = city[cur].left;
    while (cur < population && city[cur].left == val) city[cur++].left = iLeft;
  }
  cur = 0;
  std::sort(city, city + population,
            [](const Event &a, const Event &b) { return (a.right > b.right); });
  for (int iRight = 1; cur < population; iRight++) {
    int val = city[cur].right;
    while (cur < population && city[cur].right == val)
      city[cur++].right = iRight;
  }
  cur = 0;
  std::sort(city, city + population, [](const Event &a, const Event &b) {
    return (a.pos != b.pos ? a.pos < b.pos : a.type > b.type);
  });
  for (int iEvent = 0; iEvent < population; iEvent++) {
    if (city[iEvent].type == 0)
      canBuy[city[iEvent].id] = cur - sum(leftTree, city[iEvent].left) -
                                sum(rightTree, city[iEvent].right);
    else {
      modif(leftTree, city[iEvent].left, city[iEvent].type);
      modif(rightTree, city[iEvent].right, city[iEvent].type);
      cur += city[iEvent].type;
    }
  }
  for (int iPers = 0; iPers < people; iPers++)
    std::cout << canBuy[iPers] << ' ';
  std::cout << '\n';
  return 0;
}