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 N = 1e5, M = 1e5;
int n, m, p[N], s[N], b[N], inc[M], pref[M];
struct event {
  int type, pos, add, id;
};
map<int, vector<event>> mp;
set<int> s1, s2;
vector<int> v1, v2;
int f1[2 * N], f2[2 * N], ans[M];
void update1(int pos, int dif) {
  for (; pos < v1.size(); pos |= pos + 1) f1[pos] += dif;
}
int query1(int pos) {
  int res = 0;
  for (; pos > 0; pos &= pos - 1) res += f1[pos - 1];
  return res;
}
void update2(int pos, int dif) {
  for (; pos < v2.size(); pos |= pos + 1) f2[pos] += dif;
}
int query2(int pos) {
  int res = 0;
  for (; pos > 0; pos &= pos - 1) res += f2[pos - 1];
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> m;
  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];
  for (int j = 0; j < m; j++) cin >> inc[j];
  for (int j = 0; j < m; j++) cin >> pref[j];
  for (int i = 0; i < n; i++) {
    event e;
    e.type = 1;
    e.pos = p[i] + b[i];
    e.add = 1;
    mp[p[i]].push_back(e);
    e.add = -1;
    mp[s[i] + 1].push_back(e);
    e.type = 2;
    e.pos = b[i] - p[i] + 1;
    e.add = -1;
    mp[p[i]].push_back(e);
    e.add = +1;
    mp[s[i] + 1].push_back(e);
  }
  for (int j = 0; j < m; j++) {
    event e;
    e.type = 0;
    e.pos = pref[j];
    e.id = j;
    mp[inc[j]].push_back(e);
  }
  for (int i = 0; i < n; i++) {
    s1.insert(p[i] + b[i]);
    s2.insert(b[i] - p[i] + 1);
  }
  copy(s1.begin(), s1.end(), back_inserter(v1));
  copy(s2.begin(), s2.end(), back_inserter(v2));
  for (const pair<int, vector<event>>& mp_element : mp) {
    int x = mp_element.first;
    for (event e : mp_element.second) {
      if (e.type == 1)
        update1(lower_bound(v1.begin(), v1.end(), e.pos) - v1.begin(), e.add);
      else if (e.type == 2)
        update2(lower_bound(v2.begin(), v2.end(), e.pos) - v2.begin(), e.add);
      else {
        ans[e.id] =
            query1(upper_bound(v1.begin(), v1.end(), x + e.pos) - v1.begin()) +
            query2(upper_bound(v2.begin(), v2.end(), -x + e.pos) - v2.begin());
      }
    }
  }
  for (int j = 0; j < m; j++) cout << ans[j] << " ";
  cout << endl;
  return 0;
}