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 n, m, i, nr, k;
int sol[100005], aib[2][4 * 100005], w[4 * 100005], x[100005], y[100005],
    z[100005];
struct str {
  int p, t, ind, x, y;
};
str v[3 * 100005];
int cmp(str a, str b) {
  if (a.p == b.p) {
    return a.t < b.t;
  }
  return a.p < b.p;
}
int cautbin(int x) {
  int st = 1, dr = k, mid;
  while (st <= dr) {
    mid = (st + dr) / 2;
    if (w[mid] == x) {
      return mid;
    }
    if (w[mid] < x) {
      st = mid + 1;
    } else {
      dr = mid - 1;
    }
  }
}
void update(int t, int x, int val) {
  for (; x <= k; x += (x & -x)) {
    aib[t][x] += val;
  }
}
int query(int t, int x) {
  int sum = 0;
  for (; x >= 1; x -= (x & -x)) {
    sum += aib[t][x];
  }
  return sum;
}
int main() {
  cin >> n >> m;
  for (i = 1; i <= n; i++) {
    cin >> x[i];
  }
  for (i = 1; i <= n; i++) {
    cin >> y[i];
  }
  for (i = 1; i <= n; i++) {
    cin >> z[i];
  }
  for (i = 1; i <= n; i++) {
    nr += 2;
    v[nr - 1].p = x[i];
    v[nr].p = y[i];
    v[nr - 1].t = 1;
    v[nr].t = 3;
    v[nr - 1].x = v[nr].x = x[i] + z[i];
    v[nr - 1].y = v[nr].y = x[i] - z[i];
    w[nr - 1] = v[nr].x;
    w[nr] = v[nr].y;
  }
  k = nr;
  for (i = 1; i <= m; i++) {
    cin >> x[i];
  }
  for (i = 1; i <= m; i++) {
    cin >> y[i];
  }
  for (i = 1; i <= m; i++) {
    nr++;
    v[nr].p = x[i];
    v[nr].t = 2;
    v[nr].ind = i;
    v[nr].x = x[i] + y[i];
    v[nr].y = x[i] - y[i];
    w[++k] = x[i] + y[i];
    w[++k] = x[i] - y[i];
  }
  sort(w + 1, w + k + 1);
  k = 1;
  for (i = 2; i <= 2 * n + 2 * m; i++) {
    if (w[k] != w[i]) {
      w[++k] = w[i];
    }
  }
  for (i = 1; i <= nr; i++) {
    v[i].x = cautbin(v[i].x);
    v[i].y = cautbin(v[i].y);
  }
  sort(v + 1, v + nr + 1, cmp);
  for (i = 1; i <= nr; i++) {
    if (v[i].t == 1) {
      update(0, v[i].x, 1);
      update(1, v[i].y, -1);
    }
    if (v[i].t == 2) {
      sol[v[i].ind] = query(0, v[i].x) + query(1, k) - query(1, v[i].y);
    }
    if (v[i].t == 3) {
      update(0, v[i].x, -1);
      update(1, v[i].y, 1);
    }
  }
  for (i = 1; i <= m; i++) {
    cout << sol[i] << " ";
  }
}