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;
struct event {
  int x, y, u, v;
  bool t;
  event(int x, int y, int u, int v, bool t) : x(x), y(y), u(u), v(v), t(t) {}
};
struct fenwick_tree {
  vector<int> bit;
  int n;
  void init(int n = 0) {
    bit.resize(n + 1);
    this->n = n;
  }
  void update(int i, int t) {
    for (; i <= n; i += i & -i) bit[i] += t;
  }
  int query(int i) {
    int res = 0;
    for (; i > 0; i -= i & -i) res += bit[i];
    return res;
  }
};
int main() {
  cin.tie(0)->sync_with_stdio(0);
  int n, m;
  cin >> n >> m;
  map<int, int> mp;
  vector<event> ev;
  fenwick_tree tr[2];
  vector<int> cmp[2];
  vector<int> p(n + 1), s(n + 1), b(n + 1);
  vector<int> inc(m + 1), pref(m + 1);
  vector<int> ans(m + 1);
  for (int i = 1; i <= n; i++) cin >> p[i];
  for (int i = 1; i <= n; i++) cin >> s[i];
  for (int i = 1; i <= n; i++) cin >> b[i];
  for (int i = 1; i <= m; i++) cin >> inc[i];
  for (int i = 1; i <= m; i++) cin >> pref[i];
  for (int i = 1; i <= n; i++) {
    cmp[0].emplace_back(b[i] + p[i]);
    cmp[1].emplace_back(b[i] - p[i] + 1);
  }
  sort(cmp[0].begin(), cmp[0].end());
  sort(cmp[1].begin(), cmp[1].end());
  cmp[0].erase(unique(cmp[0].begin(), cmp[0].end()), cmp[0].end());
  cmp[1].erase(unique(cmp[1].begin(), cmp[1].end()), cmp[1].end());
  tr[0].init(cmp[0].size());
  tr[1].init(cmp[1].size());
  for (int i = 1; i <= n; i++) {
    ev.emplace_back(p[i], b[i] + p[i], 1, 0, 0);
    ev.emplace_back(s[i] + 1, b[i] + p[i], -1, 0, 0);
    ev.emplace_back(p[i], b[i] - p[i] + 1, -1, 1, 0);
    ev.emplace_back(s[i] + 1, b[i] - p[i] + 1, 1, 1, 0);
  }
  for (int i = 1; i <= m; i++) {
    ev.emplace_back(inc[i], pref[i] + inc[i], i, 0, 1);
    ev.emplace_back(inc[i], pref[i] - inc[i], i, 1, 1);
  }
  sort(ev.begin(), ev.end(),
       [](event p, event q) { return p.x != q.x ? p.x < q.x : p.t < q.t; });
  for (event e : ev) {
    int i =
        upper_bound(cmp[e.v].begin(), cmp[e.v].end(), e.y) - cmp[e.v].begin();
    if (e.t)
      ans[e.u] += tr[e.v].query(i);
    else
      tr[e.v].update(i, e.u);
  }
  for (int i = 1; i <= m; i++) cout << ans[i] << " \n"[i == m];
}