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>
const int N = 4e5 + 5;
int n, m, p[N], s[N], b[N], inc[N], pre[N], ans[N];
struct Data {
  int p, x, y, opt, id;
  Data(int _p, int _x, int _y, int _opt, int _id) {
    p = _p, x = _x, y = _y, opt = _opt, id = _id;
  }
  bool operator<(const Data &b) const { return p == b.p ? id < b.id : p < b.p; }
};
struct BIT {
  int n, b[N];
  void modify(int x, int v) {
    for (; x <= n; x += x & -x) b[x] += v;
  }
  int query(int x) {
    int ans = 0;
    for (; x; x ^= x & -x) ans += b[x];
    return ans;
  }
} b1, b2;
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", &pre[i]);
  std::vector<int> v;
  for (int i = 1; i <= n; i++) {
    v.push_back(b[i] + p[i]);
    v.push_back(b[i] - p[i]);
  }
  for (int i = 1; i <= m; i++) {
    v.push_back(inc[i] + pre[i]);
    v.push_back(pre[i] - inc[i] - 1);
  }
  std::sort(v.begin(), v.end());
  v.erase(std::unique(v.begin(), v.end()), v.end());
  b1.n = b2.n = v.size();
  std::vector<Data> q;
  for (int i = 1; i <= n; i++) {
    q.push_back(Data(p[i], p[i], b[i], 1, 0));
    q.push_back(Data(s[i] + 1, p[i], b[i], -1, 0));
  }
  for (int i = 1; i <= m; i++) {
    q.push_back(Data(inc[i], inc[i], pre[i], 0, i));
  }
  std::sort(q.begin(), q.end());
  for (Data t : q) {
    if (t.opt) {
      int x = std::lower_bound(v.begin(), v.end(), t.x + t.y) - v.begin() + 1;
      int y = std::lower_bound(v.begin(), v.end(), t.y - t.x) - v.begin() + 1;
      b1.modify(x, t.opt), b2.modify(y, t.opt);
    } else {
      int x = std::lower_bound(v.begin(), v.end(), t.x + t.y) - v.begin() + 1;
      int y =
          std::lower_bound(v.begin(), v.end(), t.y - t.x - 1) - v.begin() + 1;
      ans[t.id] = b1.query(x) - b2.query(y);
    }
  }
  for (int i = 1; i <= m; i++) printf("%d%c", ans[i], " \n"[i == m]);
  return 0;
}