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 long long MAX = 2e5 + 5;
long long n, m;
long long p[MAX];
long long s[MAX];
long long b[MAX];
long long inc[MAX];
long long pref[MAX];
int A[MAX], B[MAX];
vector<long long> vec;
long long bit[MAX];
long long low(long long x) { return x & -x; }
long long que(int x) {
  long long r = 0;
  for (; x; x -= low(x)) r += bit[x];
  return r;
}
void mdy(int x, long long y) {
  for (; x < MAX; x += low(x)) bit[x] += y;
}
long long ans[MAX];
pair<long long, long long> arr[MAX];
int main() {
  cin >> n >> m;
  vector<pair<long long, long long> > vv;
  for (int i = 1; i <= n; i++) {
    cin >> p[i];
    vv.emplace_back(p[i], i);
  }
  for (int i = 1; i <= n; i++) {
    cin >> s[i];
    s[i]++;
    vv.emplace_back(s[i], -i);
  }
  sort(vv.begin(), vv.end());
  for (int i = 1; i <= n; i++) {
    cin >> b[i];
    vec.emplace_back(p[i] + b[i]);
    vec.emplace_back(b[i] - p[i]);
  }
  sort(vec.begin(), vec.end());
  vec.resize(unique(vec.begin(), vec.end()) - vec.begin());
  for (int i = 1; i <= n; i++) {
    A[i] = (lower_bound(vec.begin(), vec.end(), p[i] + b[i]) - vec.begin()) + 1;
    B[i] = (lower_bound(vec.begin(), vec.end(), b[i] - p[i]) - vec.begin()) + 1;
  }
  for (int i = 1; i <= m; i++) {
    cin >> inc[i];
    arr[i] = pair<long long, long long>(inc[i], i);
  }
  sort(arr + 1, arr + m + 1);
  for (int i = 1; i <= m; i++) cin >> pref[i];
  int now = 0;
  for (int i = 1; i <= m; i++) {
    while (now < n + n && vv[now].first <= arr[i].first) {
      long long aa = abs(vv[now].second);
      mdy(A[aa], vv[now].second / aa);
      now++;
    }
    int x = (upper_bound(vec.begin(), vec.end(),
                         inc[arr[i].second] + pref[arr[i].second]) -
             vec.begin());
    ans[arr[i].second] += que(x);
  }
  memset(bit, 0, sizeof bit);
  now = 0;
  for (int i = 1; i <= m; i++) {
    while (now < n + n && vv[now].first <= arr[i].first) {
      long long aa = abs(vv[now].second);
      mdy(B[aa], vv[now].second / aa);
      now++;
    }
    int x = (upper_bound(vec.begin(), vec.end(),
                         pref[arr[i].second] - inc[arr[i].second] - 1) -
             vec.begin());
    ans[arr[i].second] -= que(x);
  }
  for (int i = 1; i <= m; i++) {
    cout << ans[i] << ' ';
  }
}