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 mx = 1e5 + 10;
int n, m, p[mx], s[mx], b[mx];
int c[mx], f[mx], is, tot, ans[mx];
int sum[mx << 2][2], a[mx << 3];
struct node {
  int v, id;
  int u;
  bool operator<(node A) const {
    if (v == A.v) return u < A.u;
    return v < A.v;
  }
} mp[mx << 2];
int getP(int x) { return lower_bound(a + 1, a + is, x) - a; }
void add(int x, int id, int v) {
  if (x != a[getP(x)])
    while (1)
      ;
  x = getP(x);
  while (x < is) {
    sum[x][id] += v;
    x += x & (-x);
  }
}
int get(int x, int id) {
  int ans = 0;
  if (x != a[getP(x)])
    while (1)
      ;
  x = getP(x);
  while (x) {
    ans += sum[x][id];
    x -= x & (-x);
  }
  return ans;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) {
    scanf("%d", p + i);
    mp[tot++] = {p[i], i, 0};
  }
  for (int i = 1; i <= n; i++) {
    scanf("%d", s + i);
    mp[tot++] = {s[i], i, 2};
  }
  for (int i = 1; i <= n; i++) {
    scanf("%d", b + i);
    a[++is] = b[i] + p[i];
    a[++is] = b[i] + 1 - p[i];
  }
  for (int i = 1; i <= m; i++) {
    scanf("%d", c + i);
    mp[tot++] = {c[i], i, 1};
  }
  for (int i = 1; i <= m; i++) {
    scanf("%d", f + i);
    a[++is] = f[i] + c[i];
    a[++is] = f[i] - c[i];
  }
  sort(mp, mp + tot);
  sort(a + 1, a + 1 + is);
  is = unique(a + 1, a + 1 + is) - a;
  for (int i = 0; i < tot; i++) {
    int id = mp[i].id;
    if (mp[i].u & 1) {
      ans[id] = get(c[id] + f[id], 0) + get(f[id] - c[id], 1);
    } else {
      add(b[id] + p[id], 0, mp[i].u ? -1 : 1);
      add(b[id] + 1 - p[id], 1, mp[i].u ? 1 : -1);
    }
  }
  for (int i = 1; i <= m; i++) printf("%d ", ans[i]);
  return 0;
}