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 n, m, alle;
int ok[100005], p[100005], s[100005], b[100005], inc[100005], pref[100005],
    ans[100005], S[2][100005 * 16];
vector<pair<int, int> > ev;
map<int, int> cmp;
int get(int node, int bas, int son, int x, int w) {
  if (bas > x) return 0;
  if (son <= x) return S[w][node];
  return get(node << 1, bas, ((bas + son) / 2), x, w) +
         get(node << 1 | 1, ((bas + son) / 2) + 1, son, x, w);
}
void up(int node, int bas, int son, int x, int val, int w) {
  if (bas > x || son < x) return;
  if (bas == x && son == x) {
    S[w][node] += val;
    return;
  }
  up(node << 1, bas, ((bas + son) / 2), x, val, w);
  up(node << 1 | 1, ((bas + son) / 2) + 1, son, x, val, w);
  S[w][node] = S[w][node << 1] + S[w][node << 1 | 1];
}
void cm(int x) { cmp[x] = 1; }
int main() {
  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);
  for (int i = 1; i <= m; i++) scanf("%d", inc + i);
  for (int i = 1; i <= m; i++) scanf("%d", pref + i);
  for (int i = 1; i <= n; i++) {
    cm(p[i] - b[i]);
    cm(p[i] + b[i]);
    ev.push_back({p[i], i - 100005});
    ev.push_back({s[i] + 1, i - 100005});
  }
  for (int i = 1; i <= m; i++) {
    cm(inc[i] - pref[i]);
    cm(inc[i] + pref[i]);
    ev.push_back({inc[i], i});
  }
  for (auto& x : cmp) x.second = ++alle;
  sort(ev.begin(), ev.end());
  fill(ok + 1, ok + 1 + n, -1);
  int cur = 0;
  for (auto& e : ev) {
    if (e.second < 0) {
      e.second += 100005;
      ok[e.second] = -ok[e.second];
      cur += ok[e.second];
      up(1, 1, alle, cmp[p[e.second] - b[e.second]], ok[e.second], 0);
      up(1, 1, alle, cmp[p[e.second] + b[e.second]], ok[e.second], 1);
    } else {
      ans[e.second] = get(1, 1, alle, cmp[inc[e.second] - pref[e.second]], 0) +
                      get(1, 1, alle, cmp[inc[e.second] + pref[e.second]], 1) -
                      cur;
    }
  }
  for (int i = 1; i <= m; i++) printf("%d ", ans[i]);
}