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 = 2 * (1e9) + 10;
const long long infl = 2 * (1e18) + 10;
const long long MOD = 1000000007;
const long NMAX = 400005;
const int dx[] = {-1, 0, 1, 0};
const int dy[] = {0, 1, 0, -1};
int lb[NMAX], rb[NMAX], yd[NMAX], qx[NMAX], qy[NMAX];
int ans[NMAX];
struct AIB {
  int t[4 * NMAX];
  int sum(int n, int l, int r, int ql, int qr) {
    if (ql <= l && r <= qr) return t[n];
    int pivot = (l + r) / 2;
    int ans = 0;
    if (ql <= pivot) ans += sum(2 * n, l, pivot, ql, qr);
    if (qr > pivot) ans += sum(2 * n + 1, pivot + 1, r, ql, qr);
    return ans;
  }
  void update(int n, int l, int r, int x, int y) {
    if (l == r) {
      t[n] += y;
      return;
    }
    int pivot = (l + r) / 2;
    if (x <= pivot)
      update(2 * n, l, pivot, x, y);
    else
      update(2 * n + 1, pivot + 1, r, x, y);
    t[n] = t[2 * n] + t[2 * n + 1];
  }
};
AIB up, down;
int main() {
  cin.sync_with_stdio(0);
  int n, m;
  cin >> n >> m;
  for (int i = 1; i <= n; ++i) cin >> lb[i];
  for (int i = 1; i <= n; ++i) cin >> rb[i];
  for (int i = 1; i <= n; ++i) cin >> yd[i];
  for (int i = 1; i <= m; ++i) cin >> qx[i];
  for (int i = 1; i <= m; ++i) cin >> qy[i];
  vector<pair<int, pair<int, int>>> axi;
  for (int i = 1; i <= n; ++i) {
    axi.push_back(make_pair(lb[i], make_pair(1, i)));
    axi.push_back(make_pair(rb[i], make_pair(3, i)));
  }
  for (int i = 1; i <= m; ++i) {
    axi.push_back(make_pair(qx[i], make_pair(2, i)));
  }
  sort(axi.begin(), axi.end());
  map<int, int> upm, downm;
  vector<int> upvalues, downvalues;
  for (int i = 1; i <= n; ++i) {
    upvalues.push_back(lb[i] - yd[i]);
    downvalues.push_back(lb[i] + yd[i]);
  }
  for (int i = 1; i <= m; ++i) {
    upvalues.push_back(qx[i] - qy[i]);
    downvalues.push_back(qx[i] + qy[i]);
  }
  sort(upvalues.begin(), upvalues.end());
  sort(downvalues.begin(), downvalues.end());
  for (int i = 0; i < n + m; ++i) {
    upm[upvalues[i]] = i + 1;
    downm[downvalues[i]] = i + 1;
  }
  for (auto el : axi) {
    int id = el.second.second;
    int typ = el.second.first;
    if (typ == 1) {
      up.update(1, 1, n + m + 1, upm[lb[id] - yd[id]], +1);
      down.update(1, 1, n + m + 1, downm[lb[id] + yd[id]], +1);
    }
    if (typ == 3) {
      up.update(1, 1, n + m + 1, upm[lb[id] - yd[id]], -1);
      down.update(1, 1, n + m + 1, downm[lb[id] + yd[id]], -1);
    }
    if (typ == 2) {
      ans[id] = down.sum(1, 1, n + m + 1, 1, downm[qx[id] + qy[id]]) -
                up.sum(1, 1, n + m + 1, upm[qx[id] - qy[id]] + 1, n + m + 1);
    }
  }
  for (int i = 1; i <= m; ++i) {
    cout << ans[i] << " ";
  }
  return 0;
}