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;
int p[100005], s[100005], b[100005], inc[100005], pref[100005];
int ans[100005];
bool vis[100005];
map<int, vector<pair<int, int> > > P;
map<int, vector<pair<int, int> > >::iterator its;
map<int, int> com;
map<int, int>::iterator its2;
struct ft {
  int ft[400005];
  void update(int i, int v) {
    for (; i <= 400000; i += (i & -i)) {
      ft[i] += v;
    }
  }
  int query(int i) {
    int res = 0;
    for (; i > 0; i -= (i & -i)) res += ft[i];
    return res;
  }
} T[2];
int main() {
  int n, m, c = 1;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", &p[i]);
  for (int i = 1; i <= n; i++) scanf("%d", &s[i]);
  for (int i = 1; i <= n; i++)
    scanf("%d", &b[i]), P[p[i]].push_back(pair<int, int>(0, i)),
        P[s[i] + 1].push_back(pair<int, int>(0, i)),
        com[p[i] + b[i]] = 1, com[-(p[i] - b[i] - 1)] = 1;
  for (int i = 1; i <= m; i++) scanf("%d", &inc[i]);
  for (int i = 1; i <= m; i++)
    scanf("%d", &pref[i]), P[inc[i]].push_back(pair<int, int>(1, i)),
        com[inc[i] + pref[i]] = 1, com[-(inc[i] - pref[i])] = 1;
  for (its2 = com.begin(); its2 != com.end(); its2++) its2->second = c++;
  for (its = P.begin(); its != P.end(); its++) {
    vector<pair<int, int> > v = its->second;
    sort(v.begin(), v.end());
    for (int i = 0; i < v.size(); i++) {
      int t = v[i].first, j = v[i].second;
      if (t == 0) {
        if (!vis[j]) {
          vis[j] = 1;
          int l = com[p[j] + b[j]];
          int r = com[-(p[j] - b[j] - 1)];
          T[0].update(l, 1);
          T[1].update(r, 1);
        } else {
          vis[j] = 0;
          int l = com[p[j] + b[j]];
          int r = com[-(p[j] - b[j] - 1)];
          T[0].update(l, -1);
          T[1].update(r, -1);
        }
      } else {
        int l = com[inc[j] + pref[j]];
        int r = com[-(inc[j] - pref[j])];
        ans[j] = T[0].query(l) - T[1].query(r);
      }
    }
  }
  for (int i = 1; i <= m; i++) printf("%d ", ans[i]);
}