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;
long long mod = 1000000007;
int n, m;
struct Node {
  long long x;
  int id;
  int flag;
  Node() {}
  Node(int x, int id, int flag) : x(x), id(id), flag(flag) {}
};
const int maxn = 1e5 + 10;
vector<Node> ft;
vector<long long> st;
map<long long, int> idx;
long long sum1[500010], sum2[500010];
int get_idx(long long t) { return idx[t]; }
void add1(long long t, int flag) {
  for (; t <= 4 * maxn; t += t & -t) {
    sum1[t] += flag;
  }
}
void add2(long long t, int flag) {
  for (; t <= 4 * maxn; t += t & -t) {
    sum2[t] += flag;
  }
}
int get1(long long t) {
  int ret = 0;
  while (t > 0) {
    ret += sum1[t];
    t -= t & -t;
  }
  return ret;
}
int get2(long long t) {
  int ret = 0;
  while (t > 0) {
    ret += sum2[t];
    t -= t & -t;
  }
  return ret;
}
int p[100010], s[100010], b[100010], inc[100010], pre[100010];
int ans[100010];
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 0; i < n; ++i) scanf("%d", &p[i]);
  for (int i = 0; i < n; ++i) scanf("%d", &s[i]);
  for (int i = 0; i < n; ++i) scanf("%d", &b[i]);
  for (int i = 0; i < m; ++i) scanf("%d", &inc[i]);
  for (int i = 0; i < m; ++i) scanf("%d", &pre[i]);
  for (int i = 0; i < n; ++i) {
    ft.push_back(Node(p[i], i, 0));
    ft.push_back(Node(s[i], i, 2));
    st.push_back(b[i] + p[i]);
    st.push_back(b[i] - p[i] + 1);
  }
  for (int i = 0; i < m; ++i) {
    ft.push_back(Node(inc[i], i, 1));
    st.push_back(pre[i] - inc[i]);
    st.push_back(inc[i] + pre[i]);
  }
  auto cmp = [&](const Node &a, const Node &b) {
    return a.x < b.x || (a.x == b.x && a.flag < b.flag);
  };
  sort(ft.begin(), ft.end(), cmp);
  sort(st.begin(), st.end());
  int pos = unique(st.begin(), st.end()) - st.begin();
  for (int i = 0; i < pos; ++i) {
    idx[st[i]] = i + 1;
  }
  memset(sum1, 0, sizeof(sum1));
  memset(sum2, 0, sizeof(sum2));
  for (int i = 0; i < ft.size(); ++i) {
    Node node = ft[i];
    int id = node.id;
    int flag = node.flag;
    if (flag == 0) {
      add1(get_idx(b[id] + p[id]), 1);
      add2(get_idx(b[id] - p[id] + 1), -1);
    } else if (flag == 2) {
      add1(get_idx(b[id] + p[id]), -1);
      add2(get_idx(b[id] - p[id] + 1), 1);
    } else {
      ans[id] =
          get1(get_idx(inc[id] + pre[id])) + get2(get_idx(pre[id] - inc[id]));
      if (ans[id] < 0) ans[id] = -ans[id];
    }
  }
  for (int i = 0; i < m; ++i) {
    printf("%d ", ans[i]);
  }
  printf("\n");
}