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 N = 1200000 + 10;
template <typename T>
class fenwick_tree {
 private:
  int n;
  vector<T> fenw;

 public:
  fenwick_tree(){};
  void init(int _n) {
    n = _n;
    fenw.resize(n, 0);
  }
  size_t size() { return n; }
  void update(int x, T val) {
    while (x < n) {
      fenw[x] += val;
      x |= (x + 1);
    }
  }
  T query(int x) {
    T ans = 0;
    while (x >= 0) {
      ans += fenw[x];
      x = (x & (x + 1)) - 1;
    }
    return ans;
  }
};
struct node {
  int x, y, z, t, type;
} arr[N];
int n, m, cnt;
int p[N], s[N], b[N];
int inc[N], pref[N], ans[N];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.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];
  set<int> st;
  for (int i = 1; i <= n; ++i) {
    st.insert(b[i] + p[i]);
    st.insert(b[i] - p[i] + 1);
  }
  for (int i = 1; i <= m; ++i) {
    st.insert(pref[i] + inc[i]);
    st.insert(pref[i] - inc[i]);
  }
  map<int, int> mp;
  int cur = 0;
  for (auto &x : st) {
    mp[x] = cur++;
  }
  for (int i = 1; i <= n; ++i) {
    arr[cnt++] = {p[i], mp[b[i] + p[i]], 1, 0, 0};
    arr[cnt++] = {p[i], mp[b[i] - p[i] + 1], -1, 1, 0};
    arr[cnt++] = {s[i] + 1, mp[b[i] + p[i]], -1, 0, 0};
    arr[cnt++] = {s[i] + 1, mp[b[i] - p[i] + 1], 1, 1, 0};
  }
  for (int i = 1; i <= m; ++i) {
    arr[cnt++] = {inc[i], mp[pref[i] + inc[i]], i, 0, 1};
    arr[cnt++] = {inc[i], mp[pref[i] - inc[i]], i, 1, 1};
  }
  sort(arr, arr + cnt, [](const node &a, const node &b) {
    if (a.x != b.x) return a.x < b.x;
    return a.type < b.type;
  });
  fenwick_tree<int> tree[2];
  tree[0].init(cur);
  tree[1].init(cur);
  for (int i = 0; i < cnt; ++i) {
    if (arr[i].type)
      ans[arr[i].z] += tree[arr[i].t].query(arr[i].y);
    else
      tree[arr[i].t].update(arr[i].y, arr[i].z);
  }
  for (int i = 1; i <= m; ++i) cout << ans[i] << ' ';
  cout << '\n';
}