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 p[300001], s[300001], b[300001], inc[300001], pref[300001], T[300001],
    ans[300001];
vector<vector<pair<int, int> > > queries(300001);
vector<int> vR;
void RemoveDuplicates(vector<int> &v) {
  sort(v.begin(), v.end());
  v.erase(unique(v.begin(), v.end()), v.end());
}
int GetCompressedValue(vector<int> &v, int x) {
  return lower_bound(v.begin(), v.end(), x) - v.begin() + 1;
}
void Update(int i, int x) {
  for (; i < 300001; i += (i & -i)) {
    T[i] += x;
  }
}
int Query(int i) {
  int ans = 0;
  for (; i > 0; i -= (i & -i)) {
    ans += T[i];
  }
  return ans;
}
void Solve() {
  vector<int> v;
  for (int i = 1; i < 300001; ++i) {
    for (auto query : queries[i]) {
      v.push_back(query.first);
    }
  }
  RemoveDuplicates(v);
  for (int j = 1; j < 300001; ++j) {
    for (auto query : queries[j]) {
      int i = GetCompressedValue(v, query.first);
      int x = query.second;
      if (x < 2) {
        Update(i, x);
      } else {
        ans[x - 1] += Query(i);
      }
    }
  }
}
void AddToQuery(int i, int x, int y) { queries[i].push_back(make_pair(x, y)); }
void SolveLeftPart(int n, int m) {
  for (int i = 1; i <= n; ++i) {
    AddToQuery(GetCompressedValue(vR, p[i]), p[i] + b[i], 1);
    AddToQuery(GetCompressedValue(vR, s[i] + 1), p[i] + b[i], -1);
  }
  for (int i = 1; i <= m; ++i) {
    AddToQuery(GetCompressedValue(vR, inc[i]), inc[i] + pref[i], i + 1);
  }
  Solve();
}
void SolveRightPart(int n, int m) {
  for (int i = 1; i < 300001; ++i) {
    T[i] = 0;
    queries[i].clear();
  }
  for (int i = 1; i <= n; ++i) {
    AddToQuery(GetCompressedValue(vR, p[i]), b[i] - p[i] + 1, -1);
    AddToQuery(GetCompressedValue(vR, s[i] + 1), b[i] - p[i] + 1, 1);
  }
  for (int i = 1; i <= m; ++i) {
    AddToQuery(GetCompressedValue(vR, inc[i]), pref[i] - inc[i], i + 1);
  }
  Solve();
}
void TakeInput(int n, int a[]) {
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
  }
}
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  TakeInput(n, p);
  TakeInput(n, s);
  TakeInput(n, b);
  TakeInput(m, inc);
  TakeInput(m, pref);
  for (int i = 1; i <= n; ++i) {
    vR.push_back(p[i]);
    vR.push_back(s[i] + 1);
  }
  for (int i = 1; i <= m; ++i) {
    vR.push_back(inc[i]);
  }
  RemoveDuplicates(vR);
  SolveLeftPart(n, m);
  SolveRightPart(n, m);
  for (int i = 1; i <= m; ++i) {
    printf("%d ", ans[i]);
  }
  printf("\n");
  return 0;
}