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 Inf = 2000000007;
const int Maxn = 100005;
vector<int> un;
int n, m;
int B[Maxn];
int mylin[Maxn];
pair<int, int> lin[Maxn];
int A[Maxn];
int Y[Maxn], X[Maxn];
vector<pair<int, int> > seq1;
vector<pair<int, int> > seq2;
vector<int> st[Maxn];
vector<int> BIT[Maxn];
int res[Maxn];
void insertIner(vector<int> &BIT, int ind, int val) {
  ind++;
  for (int i = ind; i > 0; i -= i & -i) BIT[i] += val;
}
int getIner(const vector<int> &BIT, int ind) {
  ind++;
  int res = 0;
  for (int i = ind; i < BIT.size(); i += i & -i) res += BIT[i];
  return res;
}
void attachOuter(int ind, int val) {
  ind++;
  for (int i = ind; i <= un.size(); i += i & -i)
    if (st[i].empty() || st[i].back() != val) st[i].push_back(val);
}
void createOuter() {
  for (int i = 1; i <= un.size(); i++) BIT[i].resize(st[i].size() + 1, 0);
}
void addOuter(int ind, int y, int val) {
  ind++;
  for (int i = ind; i <= un.size(); i += i & -i) {
    int ind = lower_bound(st[i].begin(), st[i].end(), y) - st[i].begin();
    insertIner(BIT[i], ind, val);
  }
}
int getOuter(int ind, int y) {
  ind++;
  int res = 0;
  for (int i = ind; i > 0; i -= i & -i) {
    int ind = lower_bound(st[i].begin(), st[i].end(), y) - st[i].begin();
    res += getIner(BIT[i], ind);
  }
  return res;
}
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 0; i < n; i++) scanf("%d", &B[i]);
  for (int i = 0; i < n; i++) {
    scanf("%d", &lin[i].first);
    lin[i].second = i;
    mylin[i] = lin[i].first;
  }
  for (int i = 0; i < n; i++) {
    scanf("%d", &A[i]);
    un.push_back(B[i] - A[i]);
    seq1.push_back(pair<int, int>(A[i] + B[i], i));
  }
  sort(seq1.begin(), seq1.end());
  sort(un.begin(), un.end());
  un.erase(unique(un.begin(), un.end()), un.end());
  sort(lin, lin + n);
  for (int i = 0; i < n; i++) {
    int ind = lin[i].second;
    int bitind =
        lower_bound(un.begin(), un.end(), B[ind] - A[ind]) - un.begin();
    attachOuter(bitind, lin[i].first);
  }
  createOuter();
  for (int i = 0; i < m; i++) scanf("%d", &Y[i]);
  for (int i = 0; i < m; i++) {
    scanf("%d", &X[i]);
    seq2.push_back(pair<int, int>(X[i] + Y[i], i));
  }
  sort(seq2.begin(), seq2.end());
  int pnt = 0;
  for (int i = 0; i < seq2.size(); i++) {
    while (pnt < seq1.size() && seq1[pnt].first <= seq2[i].first) {
      int ind = seq1[pnt].second;
      int bitind =
          lower_bound(un.begin(), un.end(), B[ind] - A[ind]) - un.begin();
      addOuter(bitind, mylin[ind], 1);
      pnt++;
    }
    int ind = seq2[i].second;
    int bitind =
        upper_bound(un.begin(), un.end(), Y[ind] - X[ind]) - un.begin() - 1;
    res[ind] = getOuter(bitind, Y[ind]);
  }
  for (int i = 0; i < m; i++) printf("%d%c", res[i], i + 1 < m ? ' ' : '\n');
  return 0;
}