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;
const int MXN = 1e5 + 5;
int p[MXN];
int s[MXN];
int b[MXN];
int inc[MXN];
int srt[MXN];
int sol[MXN];
int pref[MXN];
array<int, 3> dish[MXN];
array<int, 2> person[MXN];
struct fenwick {
  int ft[MXN];
  void update(int pos, int dx) {
    for (int x = pos + 1; x < MXN; x += x & -x) {
      ft[x] += dx;
    }
  }
  int query(int pos) {
    int ret = 0;
    for (int x = pos + 1; x; x -= x & -x) {
      ret += ft[x];
    }
    return ret;
  }
} fen[2];
int main() {
  ios::sync_with_stdio(0), cin.tie(0);
  int n, m;
  cin >> n >> m;
  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<array<int, 3>> sweep[2];
  for (int i = 0; i < n; ++i) {
    sweep[0].push_back({b[i] + p[i], 0, i});
    sweep[1].push_back({b[i] - p[i], 1, i});
  }
  for (int i = 0; i < m; ++i) {
    sweep[0].push_back({pref[i] + inc[i], 1, i});
    sweep[1].push_back({pref[i] - inc[i], 0, i});
  }
  for (int i = 0; i < m; ++i) {
    srt[i] = inc[i];
  }
  sort(srt, srt + m);
  sort(begin(sweep[0]), end(sweep[0]));
  sort(begin(sweep[1]), end(sweep[1]));
  for (int i = 0; i < n + m; ++i) {
    for (int j = 0; j < 2; ++j) {
      int t = sweep[j][i][1];
      int x = sweep[j][i][2];
      if (t == j) {
        fen[j].update(lower_bound(srt, srt + m, p[x]) - srt, +1);
        fen[j].update(upper_bound(srt, srt + m, s[x]) - srt, -1);
      } else {
        int pos = lower_bound(srt, srt + m, inc[x]) - srt;
        sol[x] += (j ? -1 : 1) * fen[j].query(pos);
      }
    }
  }
  for (int i = 0; i < m; ++i) {
    cout << sol[i] << ' ';
  }
}