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 const maxn = 1e5 + 5;
int p[maxn], s[maxn], b[maxn], inc[maxn], pref[maxn], go[maxn], ptr[(1 << 20)],
    ans[maxn], f1[maxn], f2[maxn], t[2 * maxn];
vector<int> Q;
vector<int> tree[(1 << 20)];
vector<pair<int, int> > ask_pref[(1 << 20)], ask_suff[(1 << 20)];
int N;
inline void add(int i, int x) {
  while (i <= N) {
    t[i] += x;
    i = (i | (i + 1));
  }
}
inline int sum(int i) {
  int ans = 0;
  while (i >= 0) {
    ans += t[i];
    i = (i & (i + 1)) - 1;
  }
  return ans;
}
inline int lower(int x) {
  return lower_bound(Q.begin(), Q.end(), x) - Q.begin();
}
inline int upper(int x) {
  return upper_bound(Q.begin(), Q.end(), x) - Q.begin();
}
void update(int i, int l, int r, int lq, int rq, int x) {
  if (lq >= r || l >= rq) return;
  if (lq <= l && r <= rq) {
    tree[i].push_back(x);
    return;
  }
  int m = (r + l) / 2;
  update(2 * i + 1, l, m, lq, rq, x);
  update(2 * i + 2, m, r, lq, rq, x);
}
void get(int i, int l, int r, int lq, int x) {
  while (ptr[i] < (int)tree[i].size() && pref[x] > b[tree[i][ptr[i]]]) ptr[i]++;
  ask_pref[i].push_back({ptr[i] - 1, x});
  ask_suff[i].push_back({ptr[i], x});
  if (r - l == 1) return;
  int m = (r + l) / 2;
  if (lq < m)
    get(2 * i + 1, l, m, lq, x);
  else
    get(2 * i + 2, m, r, lq, x);
}
void go1(int i, int l, int r) {
  int ptr = 0;
  for (int j = 0; j < (int)tree[i].size(); ++j) {
    while (ptr < (int)ask_pref[i].size() && ask_pref[i][ptr].first < j) {
      ans[ask_pref[i][ptr].second] += sum(f2[ask_pref[i][ptr].second]);
      ptr++;
    }
    add(f1[tree[i][j]], 1);
  }
  while (ptr < (int)ask_pref[i].size()) {
    ans[ask_pref[i][ptr].second] += sum(f2[ask_pref[i][ptr].second]);
    ptr++;
  }
  for (int j = 0; j < (int)tree[i].size(); ++j) add(f1[tree[i][j]], -1);
  if (r - l == 1) return;
  int m = (r + l) / 2;
  go1(2 * i + 1, l, m);
  go1(2 * i + 2, m, r);
}
void go2(int i, int l, int r) {
  int ptr = (int)ask_suff[i].size() - 1;
  for (int j = (int)tree[i].size() - 1; j >= 0; --j) {
    while (ptr >= 0 && ask_suff[i][ptr].first > j) {
      ans[ask_suff[i][ptr].second] += sum(f2[ask_suff[i][ptr].second]);
      ptr--;
    }
    add(f1[tree[i][j]], 1);
  }
  while (ptr >= 0) {
    ans[ask_suff[i][ptr].second] += sum(f2[ask_suff[i][ptr].second]);
    ptr--;
  }
  for (auto key : tree[i]) add(f1[key], -1);
  if (r - l == 1) return;
  int m = (r + l) / 2;
  go2(2 * i + 1, l, m);
  go2(2 * i + 2, m, r);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, m;
  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) go[i] = i;
  for (int i = 1; i <= n; ++i) {
    Q.push_back(p[i]), Q.push_back(s[i]);
  }
  for (int i = 1; i <= m; ++i) Q.push_back(inc[i]);
  sort(Q.begin(), Q.end());
  sort(go + 1, go + n + 1, [](int x, int y) { return b[x] < b[y]; });
  for (int i = 1; i <= n; ++i)
    update(0, 0, (int)Q.size(), lower(p[go[i]]), upper(s[go[i]]), go[i]);
  for (int i = 1; i <= m; ++i) go[i] = i;
  sort(go + 1, go + m + 1, [](int x, int y) { return pref[x] < pref[y]; });
  for (int i = 1; i <= m; ++i) {
    get(0, 0, (int)Q.size(), lower(inc[go[i]]), go[i]);
  }
  int MAXN = (int)Q.size();
  Q = {};
  for (int i = 1; i <= n; ++i) Q.push_back(b[i] + p[i]);
  for (int i = 1; i <= m; ++i) Q.push_back(inc[i] + pref[i]);
  sort(Q.begin(), Q.end());
  for (int i = 1; i <= n; ++i) f1[i] = lower(b[i] + p[i]);
  for (int i = 1; i <= m; ++i) f2[i] = lower(inc[i] + pref[i]);
  N = n + m;
  go2(0, 0, MAXN);
  Q = {};
  for (int i = 1; i <= n; ++i) Q.push_back(p[i] - b[i]);
  for (int i = 1; i <= m; ++i) Q.push_back(inc[i] - pref[i]);
  sort(Q.begin(), Q.end());
  for (int i = 1; i <= n; ++i) f1[i] = lower(p[i] - b[i]);
  for (int i = 1; i <= m; ++i) f2[i] = lower(inc[i] - pref[i]);
  go1(0, 0, MAXN);
  for (int i = 1; i <= m; ++i) cout << ans[i] << " ";
  return 0;
}