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 = 1e5 + 20;
int n, m;
int p[N], s[N], b[N];
int inc[N], pref[N];
int sz;
struct node {
  vector<int> v1, v2;
} seg[(5 * N) << 2];
vector<int> vec;
void compress() {
  for (int i = 0; i < n; i++) {
    vec.push_back(p[i]);
    vec.push_back(s[i]);
    vec.push_back(b[i]);
  }
  for (int i = 0; i < m; i++) {
    vec.push_back(inc[i]);
    vec.push_back(pref[i]);
  }
  sort(vec.begin(), vec.end());
  vec.resize(unique(vec.begin(), vec.end()) - vec.begin());
  sz = (int)vec.size();
  for (int i = 0; i < n; i++) {
    p[i] = lower_bound(vec.begin(), vec.end(), p[i]) - vec.begin();
    s[i] = lower_bound(vec.begin(), vec.end(), s[i]) - vec.begin();
    b[i] = lower_bound(vec.begin(), vec.end(), b[i]) - vec.begin();
  }
  for (int i = 0; i < m; i++) {
    inc[i] = lower_bound(vec.begin(), vec.end(), inc[i]) - vec.begin();
    pref[i] = lower_bound(vec.begin(), vec.end(), pref[i]) - vec.begin();
  }
}
void add(int l, int r, int val1, int val2, int v = 1, int s = 0, int e = sz) {
  if (r <= s || e <= l) return;
  if (l <= s && e <= r) {
    seg[v].v1.push_back(val1);
    seg[v].v2.push_back(val2);
    return;
  }
  add(l, r, val1, val2, (v << 1), s, (s + e) >> 1);
  add(l, r, val1, val2, ((v << 1) ^ 1), (s + e) >> 1, e);
}
void go(int v = 1, int s = 0, int e = sz) {
  sort(seg[v].v1.begin(), seg[v].v1.end());
  sort(seg[v].v2.begin(), seg[v].v2.end());
  if (e - s == 1) return;
  go((v << 1), s, (s + e) >> 1);
  go(((v << 1) ^ 1), (s + e) >> 1, e);
}
int get(int a, int b, int v = 1, int s = 0, int e = sz) {
  int sum = 0;
  sum += (int)seg[v].v1.size();
  sum -= seg[v].v1.end() -
         upper_bound(seg[v].v1.begin(), seg[v].v1.end(), vec[b] + vec[a]);
  sum -= lower_bound(seg[v].v2.begin(), seg[v].v2.end(), vec[b] - vec[a]) -
         seg[v].v2.begin();
  if (e - s == 1) return sum;
  if (a < (s + e) >> 1)
    sum += get(a, b, (v << 1), s, (s + e) >> 1);
  else
    sum += get(a, b, ((v << 1) ^ 1), (s + e) >> 1, e);
  return sum;
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m;
  for (int i = 0; i < n; i++) cin >> p[i];
  for (int i = 0; i < n; i++) cin >> s[i], s[i]++;
  for (int i = 0; i < n; i++) cin >> b[i];
  for (int i = 0; i < m; i++) cin >> inc[i];
  for (int i = 0; i < m; i++) cin >> pref[i];
  compress();
  for (int i = 0; i < n; i++)
    add(p[i], s[i], vec[b[i]] + vec[p[i]], vec[b[i]] - vec[p[i]]);
  go();
  for (int i = 0; i < m; i++) {
    cout << get(inc[i], pref[i]) << ' ';
  }
  cout << '\n';
  return 0;
}