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 maxn = 1e5 + 10;
struct Node {
  int is, id, x;
  bool operator<(const Node &a) const {
    if (x != a.x)
      return x < a.x;
    else
      return is < a.is;
  }
};
int a[maxn * 4], s[maxn], p[maxn], inc[maxn], b[maxn], treea[maxn * 4],
    treeb[maxn * 4], pb[maxn];
int ans[maxn];
map<int, int> ma;
int cnt = 0, tt = 0;
Node que[maxn * 3];
int getid(int x) {
  x = -x;
  return ma[x];
}
void add1(int x, int y) {
  for (int i = x; i < 4 * maxn; i += (i & -i)) treea[i] += y;
}
void add2(int x, int y) {
  for (int i = x; i < 4 * maxn; i += (i & -i)) treeb[i] += -y;
}
int quer1(int x) {
  int res = 0;
  for (int i = x; i >= 1; i -= (i & -i)) res += treea[i];
  return res;
}
int quer2(int x) {
  int res = 0;
  for (int i = x; i >= 1; i -= (i & -i)) res += treeb[i];
  return res;
}
int main() {
  int n, m;
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", &p[i]);
  for (int i = 1; i <= n; i++) scanf("%d", &s[i]);
  for (int i = 1; i <= n; i++) scanf("%d", &b[i]);
  for (int i = 1; i <= m; i++) scanf("%d", &inc[i]);
  for (int i = 1; i <= m; i++) scanf("%d", &pb[i]);
  for (int i = 1; i <= n; i++) {
    a[++cnt] = -(p[i] + b[i] - 1), a[++cnt] = -(b[i] - p[i]);
    que[++tt] = Node{0, i, p[i]};
    que[++tt] = Node{2, i, s[i]};
  }
  for (int i = 1; i <= m; i++) {
    a[++cnt] = -(inc[i] + pb[i]), a[++cnt] = -(pb[i] - inc[i]);
    que[++tt] = Node{1, i, inc[i]};
  }
  sort(que + 1, que + 1 + tt);
  sort(a + 1, a + cnt + 1);
  int pz = 1;
  for (int i = 1; i <= cnt; i++)
    if (ma[a[i]] == 0) ma[a[i]] = pz++;
  for (int i = 1; i <= tt; i++) {
    Node now = que[i];
    if (now.is == 0) {
      add1(getid(b[now.id] - p[now.id]), 1);
      add2(getid(p[now.id] + b[now.id] - 1), 1);
    } else if (now.is == 1) {
      ans[now.id] = quer1(getid(pb[now.id] - inc[now.id])) +
                    quer2(getid(inc[now.id] + pb[now.id]));
    } else if (now.is == 2) {
      add1(getid(b[now.id] - p[now.id]), -1);
      add2(getid(p[now.id] + b[now.id] - 1), -1);
    }
  }
  for (int i = 1; i <= m; i++) {
    printf("%d ", ans[i]);
  }
  return 0;
}