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;
void debug_out() { cerr << endl; }
template <class T>
ostream& prnt(ostream& out, T v) {
  out << v.size() << '\n';
  for (auto e : v) out << e << ' ';
  return out;
}
template <class T>
ostream& operator<<(ostream& out, vector<T> v) {
  return prnt(out, v);
}
template <class T>
ostream& operator<<(ostream& out, set<T> v) {
  return prnt(out, v);
}
template <class T1, class T2>
ostream& operator<<(ostream& out, map<T1, T2> v) {
  return prnt(out, v);
}
template <class T1, class T2>
ostream& operator<<(ostream& out, pair<T1, T2> p) {
  return out << '(' << p.first << ' ' << p.second << ')';
}
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  cerr << " " << H;
  debug_out(T...);
}
long long n, m;
const long long N = 100100;
const long long MAX = 4e9 + 1000;
const long long OFF = 1e9;
unordered_map<long long, long long> l, r;
long long p[N], s[N], x[N], y[N], b[N], ans[N];
vector<pair<pair<long long, long long>, long long> > events;
long long lsb(long long k) { return k & (-k); }
void upd(unordered_map<long long, long long>& aib, long long pos,
         long long val) {
  for (; pos < MAX; pos += lsb(pos)) aib[pos] += val;
}
long long que(unordered_map<long long, long long>& aib, long long pos) {
  long long ret = 0;
  for (; pos; pos -= lsb(pos)) ret += aib[pos];
  return ret;
}
long long query(long long x, long long y) {
  return que(l, x + y) + que(r, OFF + y - x);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> m;
  for (long long i = 1; i <= n; i++) cin >> p[i];
  for (long long i = 1; i <= n; i++) cin >> s[i];
  for (long long i = 1; i <= n; i++) cin >> b[i];
  for (long long i = 1; i <= m; i++) cin >> x[i];
  for (long long i = 1; i <= m; i++) cin >> y[i];
  for (long long i = 1; i <= n; i++) {
    events.push_back({{p[i], 1}, b[i]});
    events.push_back({{s[i], 3}, b[i] + p[i]});
    events.push_back({{s[i], 4}, b[i] - p[i]});
  }
  for (long long i = 1; i <= m; i++) {
    events.push_back({{x[i], 2}, i});
  }
  sort(events.begin(), events.end());
  for (auto i : events) {
    if (i.first.second == 2)
      ans[i.second] = query(x[i.second], y[i.second]);
    else if (i.first.second == 1) {
      upd(l, i.first.first + i.second, 1);
      upd(r, OFF - i.first.first + i.second + 1, -1);
    } else if (i.first.second == 3) {
      upd(l, i.second, -1);
    } else if (i.first.second == 4) {
      upd(r, OFF + i.second + 1, 1);
    }
  }
  for (long long i = 1; i <= m; i++) cout << ans[i] << '\n';
}