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 maxn = 2e5 + 5;
int p[maxn], b[maxn], s[maxn], in[maxn], pr[maxn];
int ip[maxn], is[maxn], inp[maxn];
vector<int> v;
vector<int> A[maxn * 4], B[maxn * 4];
int n, m, t;
void compress() {
  sort(v.begin(), v.end());
  v.resize(distance(v.begin(), unique(v.begin(), v.end())));
  for (int i = 0; i < n; i++) {
    ip[i] = lower_bound(v.begin(), v.end(), p[i]) - v.begin();
    is[i] = lower_bound(v.begin(), v.end(), s[i]) - v.begin();
  }
  for (int i = 0; i < m; i++) {
    inp[i] = lower_bound(v.begin(), v.end(), in[i]) - v.begin();
  }
}
void add(int l, int r, int L, int R, int v1, int v2, int x) {
  if (l >= R || r <= L) {
    return;
  }
  if (l >= L && r <= R) {
    A[x].push_back(v1);
    B[x].push_back(v2);
    return;
  }
  int mid = (l + r) / 2;
  add(l, mid, L, R, v1, v2, x * 2);
  add(mid, r, L, R, v1, v2, x * 2 + 1);
}
int get(int l, int r, int ind, int v1, int v2, int x) {
  int tot = A[x].size();
  tot -=
      A[x].size() - (upper_bound(A[x].begin(), A[x].end(), v1) - A[x].begin());
  tot -= lower_bound(B[x].begin(), B[x].end(), v2) - B[x].begin();
  if (l + 1 >= r) {
    return tot;
  }
  int mid = (l + r) / 2;
  if (ind < mid) {
    return tot + get(l, mid, ind, v1, v2, x * 2);
  } else {
    return tot + get(mid, r, ind, v1, v2, x * 2 + 1);
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n >> m;
  for (int i = 0; i < n; i++) {
    cin >> p[i];
    v.push_back(p[i]);
  }
  for (int i = 0; i < n; i++) {
    cin >> s[i];
    v.push_back(s[i]);
  }
  for (int i = 0; i < n; i++) {
    cin >> b[i];
  }
  for (int i = 0; i < m; i++) {
    cin >> in[i];
    v.push_back(in[i]);
  }
  for (int i = 0; i < m; i++) {
    cin >> pr[i];
  }
  compress();
  t = v.size();
  for (int i = 0; i < n; i++) {
    add(0, t, ip[i], is[i] + 1, b[i] + p[i], b[i] - p[i], 1);
  }
  for (int i = 1; i < maxn * 4 - 2; i++) {
    sort(A[i].begin(), A[i].end());
    sort(B[i].begin(), B[i].end());
  }
  for (int i = 0; i < m; i++) {
    int ans = get(0, t, inp[i], pr[i] + in[i], pr[i] - in[i], 1);
    cout << ans << " ";
  }
  return 0;
}