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 = (int)1e5 + 50;
struct event {
  int type;
  int i, j, k;
};
struct BIT {
  int n;
  vector<int> arr;
  vector<int> bit;
  void init() {
    sort(arr.begin(), arr.end());
    arr.resize(unique(arr.begin(), arr.end()) - arr.begin());
    bit.resize(arr.size(), 0);
    n = bit.size();
  }
  void add(int x, int val) {
    if (x < 0) return;
    for (int i = x; i < n; i |= i + 1) bit[i] += val;
  }
  int get(int x) {
    int res = 0;
    x = min(x, n - 1);
    for (int i = x; i >= 0; i = (i & (i + 1)) - 1) res += bit[i];
    return res;
  }
} bit[2];
int n, m;
int p[N], s[N], b[N];
int inc[N], pref[N];
vector<event> evt;
int res[N];
void read(int r[], int k) {
  for (int i = 0; i < k; i++) cin >> r[i];
}
bool cmp(event &e1, event &e2) {
  if (e1.i != e2.i) return e1.i < e2.i;
  return e1.type < e2.type;
}
int lb(int i, int x) {
  return (lower_bound(bit[i].arr.begin(), bit[i].arr.end(), x) -
          bit[i].arr.begin());
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> m;
  read(p, n), read(s, n), read(b, n);
  for (int i = 0; i < n; i++) {
    evt.push_back({0, p[i], b[i] + p[i], 1});
    evt.push_back({0, s[i] + 1, b[i] + p[i], -1});
    evt.push_back({1, p[i], b[i] + 1 - p[i], -1});
    evt.push_back({1, s[i] + 1, b[i] + 1 - p[i], 1});
  }
  read(inc, m), read(pref, m);
  for (int i = 0; i < m; i++) {
    evt.push_back({2, inc[i], pref[i], i});
    bit[0].arr.push_back({pref[i] + inc[i]});
    bit[1].arr.push_back({pref[i] - inc[i]});
  }
  for (int i = 0; i < 2; i++) bit[i].init();
  sort(evt.begin(), evt.end(), cmp);
  for (const auto &e : evt) {
    if (e.type == 2) {
      res[e.k] = bit[0].get(lb(0, e.i + e.j)) + bit[1].get(lb(1, e.j - e.i));
    } else if (e.type == 0) {
      bit[0].add(lb(0, e.j), e.k);
    } else
      bit[1].add(lb(1, e.j), e.k);
  }
  for (int i = 0; i < m; i++) {
    cout << res[i] << " ";
  }
}