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;
struct fenwick1D {
  vector<int> comp;
  vector<int> fen;
  int find(int x) const {
    return lower_bound(comp.begin(), comp.end(), x + 1) - comp.begin();
  }
  void add(int x) { comp.push_back(x); }
  void init() {
    sort(comp.begin(), comp.end());
    comp.erase(unique(comp.begin(), comp.end()), comp.end());
    fen.resize(comp.size(), 0);
  }
  void update(int i, int x) {
    i = find(i);
    while (i <= comp.size()) {
      fen[i - 1] += x;
      i += i & -i;
    }
  }
  int query(int i) const {
    i = find(i);
    int ret = 0;
    while (i) {
      ret += fen[i - 1];
      i -= i & -i;
    }
    return ret;
  }
};
struct fenwick2D {
  vector<int> comp;
  vector<fenwick1D> fen;
  int find(int x) const {
    return lower_bound(comp.begin(), comp.end(), x + 1) - comp.begin();
  }
  void add1(int x) { comp.push_back(x); }
  void init1() {
    sort(comp.begin(), comp.end());
    comp.erase(unique(comp.begin(), comp.end()), comp.end());
    fen.resize(comp.size());
  }
  void add2(int i, int y) {
    i = find(i);
    while (i <= comp.size()) {
      fen[i - 1].add(y);
      i += i & -i;
    }
  }
  void init2() {
    for (auto &i : fen) i.init();
  }
  void update(int i, int j, int x) {
    i = find(i);
    while (i <= comp.size()) {
      fen[i - 1].update(j, x);
      i += i & -i;
    }
  }
  int query(int i, int j) const {
    i = find(i);
    int ret = 0;
    while (i) {
      ret += fen[i - 1].query(j);
      i -= i & -i;
    }
    return ret;
  }
} fen;
int n, m;
int P[100001];
int S[100001];
int B[100001];
int inc[100001];
int pref[100001];
int ans[100001];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m;
  for (int i = 1; i <= n; ++i) cin >> P[i];
  for (int i = 1; i <= n; ++i) cin >> S[i];
  for (int i = 1; i <= n; ++i) cin >> B[i];
  for (int i = 1; i <= m; ++i) cin >> inc[i];
  for (int i = 1; i <= m; ++i) cin >> pref[i];
  for (int i = 1; i <= n; ++i) {
    fen.add1(P[i] + B[i]);
  }
  fen.init1();
  for (int i = 1; i <= n; ++i) {
    fen.add2(P[i] + B[i], P[i] - B[i]);
  }
  fen.init2();
  vector<int> pord, nord, mord;
  for (int i = 1; i <= n; ++i) pord.push_back(i), nord.push_back(i);
  for (int i = 1; i <= m; ++i) mord.push_back(i);
  sort(pord.begin(), pord.end(), [&](int x, int y) { return P[x] < P[y]; });
  sort(nord.begin(), nord.end(), [&](int x, int y) { return S[x] < S[y]; });
  sort(mord.begin(), mord.end(), [&](int x, int y) { return inc[x] < inc[y]; });
  int j = 0, k = 0;
  for (int i : mord) {
    while (j < n && P[pord[j]] <= inc[i]) {
      int p = pord[j];
      fen.update(P[p] + B[p], P[p] - B[p], 1);
      ++j;
    }
    while (k < n && S[nord[k]] < inc[i]) {
      int p = nord[k];
      fen.update(P[p] + B[p], P[p] - B[p], -1);
      ++k;
    }
    ans[i] = fen.query(inc[i] + pref[i], inc[i] - pref[i]);
  }
  for (int i = 1; i <= m; ++i) printf("%d ", ans[i]);
  return 0;
}