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 event {
  int type, pos, x, val;
  event(int a, int b, int c, int d) {
    type = a;
    pos = b;
    x = c;
    val = d;
  }
  void Print() { cout << type << ' ' << pos << ' ' << x << ' ' << val << endl; }
};
bool comp(event A, event B) {
  if (A.pos < B.pos) return true;
  if (B.pos < A.pos) return false;
  if (A.type < B.type) return true;
  if (B.type < A.type) return false;
  if (A.x < B.x) return true;
  return false;
}
void update(int bit[], int n, int x, int val) {
  for (; x < n; x += x & -x) bit[x] += val;
}
int sum(int bit[], int x) {
  int sum = 0;
  for (; x > 0; x -= x & -x) sum += bit[x];
  return sum;
}
int main() {
  int n, m;
  cin >> n >> m;
  int p[n + 1], s[n + 1], b[n + 1], inc[m + 1], pref[m + 1];
  for (int i = 0; i < n; i++) cin >> p[i];
  for (int i = 0; i < n; i++) cin >> s[i];
  for (int i = 0; i < n; i++) cin >> b[i];
  vector<event> E;
  for (int i = 0; i < n; i++) {
    E.push_back(event(0, p[i], p[i] + b[i], 1));
    E.push_back(event(1, p[i], b[i] - p[i] + 1, -1));
    E.push_back(event(0, s[i] + 1, p[i] + b[i], -1));
    E.push_back(event(1, s[i] + 1, b[i] - p[i] + 1, 1));
  }
  for (int i = 0; i < m; i++) cin >> inc[i];
  for (int i = 0; i < m; i++) cin >> pref[i];
  map<int, int> mml, mmr;
  for (int i = 0; i < m; i++) {
    E.push_back(event(2, inc[i], pref[i], i));
    mml[inc[i] + pref[i]] = 1;
    mmr[pref[i] - inc[i]] = 1;
  }
  int num = 1;
  for (pair<int, int> x : mml) mml[x.first] = num++;
  num = 1;
  for (pair<int, int> x : mmr) mmr[x.first] = num++;
  sort(E.begin(), E.end(), comp);
  int lsize = mml.size();
  int rsize = mmr.size();
  int *lbit = new int[lsize + 1];
  int *rbit = new int[rsize + 1];
  int *ans = new int[m];
  for (int i = 0; i < lsize + 1; i++) lbit[i] = 0;
  for (int i = 0; i < rsize + 1; i++) rbit[i] = 0;
  for (int i = 0; i < E.size(); i++) {
    event e = E[i];
    if (e.type == 0) {
      auto it = mml.lower_bound(e.x);
      if (it == mml.end()) continue;
      update(lbit, lsize + 1, it->second, e.val);
    } else if (e.type == 1) {
      auto it = mmr.lower_bound(e.x);
      if (it == mmr.end()) continue;
      update(rbit, rsize + 1, it->second, e.val);
    } else {
      int lind = mml[e.x + e.pos];
      int rind = mmr[e.x - e.pos];
      int ind = e.val;
      ans[ind] = sum(lbit, lind) + sum(rbit, rind);
    }
  }
  for (int i = 0; i < m; i++) cout << ans[i] << ' ';
  cout << endl;
}