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;
class fenwick {
  int n;
  vector<long long> v;

 public:
  long long sum(int r) {
    long long rval = 0;
    for (; r >= 0; r = (r & (r + 1)) - 1) {
      rval += v[r];
    }
    return rval;
  }
  void update(int in, int d) {
    for (; in < n; in = (in | (in + 1))) {
      v[in] += d;
    }
  }
  fenwick(int N) : n(N) { v.resize(n, 0); }
};
class udalost {
 public:
  int typ, wh, i, t;
};
bool cmp(udalost& a, udalost& b) {
  if (a.t == b.t) {
    return a.typ < b.typ;
  }
  return a.t < b.t;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, m;
  cin >> n >> m;
  vector<int> p(n), s(n), b(n), inc(m), pref(m), ans(m, 0);
  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 i = 0; i < m; i++) {
    cin >> inc[i];
  }
  for (int i = 0; i < m; i++) {
    cin >> pref[i];
  }
  vector<udalost> u;
  set<long long> x;
  for (int i = 0; i < n; i++) {
    u.push_back({0, p[i] + b[i], 1, p[i]});
    u.push_back({1, b[i] + 1 - p[i], -1, p[i]});
    u.push_back({0, p[i] + b[i], -1, s[i] + 1});
    u.push_back({1, b[i] + 1 - p[i], 1, s[i] + 1});
    x.insert(p[i] + b[i]);
    x.insert(b[i] + 1 - p[i]);
  }
  for (int i = 0; i < m; i++) {
    u.push_back({2, pref[i], i, inc[i]});
    x.insert(pref[i] + inc[i]);
    x.insert(pref[i] - inc[i]);
  }
  sort(u.begin(), u.end(), cmp);
  map<int, int> xs;
  while (!x.empty()) {
    xs[*x.begin()] = xs.size();
    x.erase(x.begin());
  }
  fenwick l(xs.size()), r(xs.size());
  for (int i = 0; i < u.size(); i++) {
    if (u[i].typ == 0) {
      l.update(xs[u[i].wh], u[i].i);
    } else if (u[i].typ == 1) {
      r.update(xs[u[i].wh], u[i].i);
    } else {
      ans[u[i].i] = l.sum(xs[u[i].wh + u[i].t]) + r.sum(xs[u[i].wh - u[i].t]);
    }
  }
  for (int i = 0; i < m; i++) {
    cout << ans[i] << " ";
  }
  cout << "\n";
  return 0;
}