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 + 5;
map<int, vector<pair<int, int>>> mp;
int n, m, k, p[N], s[N], b[N], inc[N], pref[N], a[4 * N], fen[4 * N][2], ans[N];
int get_id(int x) { return lower_bound(a, a + k, x) - a; }
void add(int p, int id, int x) {
  for (p = k - get_id(p); p; p -= p & -p) fen[p][id] += x;
}
int get(int p, int id) {
  int ans = 0;
  for (p = max(k - get_id(p), 1); p <= k; p += p & -p) ans += fen[p][id];
  return ans;
}
int main() {
  cin >> n >> m;
  for (int i = 0; i < n; i++) {
    cin >> p[i];
    mp[p[i]].push_back({0, i});
  }
  for (int i = 0; i < n; i++) {
    cin >> s[i];
    mp[s[i]].push_back({2, i});
  }
  for (int i = 0; i < n; i++) {
    cin >> b[i];
    a[k++] = b[i] + s[i];
    a[k++] = b[i] - s[i];
  }
  for (int i = 0; i < m; i++) {
    cin >> inc[i];
    mp[inc[i]].push_back({1, i});
  }
  for (int i = 0; i < m; i++) {
    cin >> pref[i];
    a[k++] = pref[i] + inc[i];
    a[k++] = pref[i] - inc[i];
  }
  sort(a, a + k);
  k = unique(a, a + k) - a;
  for (auto v : mp) {
    sort(v.second.begin(), v.second.end());
    for (auto x : v.second) {
      int id = x.second;
      if (x.first & 1)
        ans[id] = get(pref[id] + inc[id], 0) + get(pref[id] - inc[id], 1);
      else
        add(b[id] + p[id], 0, x.first ? -1 : 1),
            add(b[id] - p[id] + 1, 1, x.first ? 1 : -1);
    }
  }
  for (int i = 0; i < m; i++) cout << ans[i] << ' ';
}