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;
template <typename T>
void read(T &x) {
  x = 0;
  T s = 1, c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') s = -1;
  for (; isdigit(c); c = getchar()) x = x * 10 + c - 48;
  x *= s;
}
const int maxn = 1e5 + 5;
int n, m, tot;
long long p[maxn], s[maxn], b[maxn], inc[maxn], pref[maxn];
int ans[maxn];
vector<long long> A, B;
vector<pair<long long, int>> que;
class FenwickTree {
 public:
  int F[maxn << 1];
  void Modify(int x, int val) {
    for (; x <= maxn << 1; x += x & -x) F[x] += val;
  }
  int Query(int x) {
    int ret = 0;
    for (; x; x -= x & -x) ret += F[x];
    return ret;
  }
} bitA, bitB;
inline void Read() {
  read(n), read(m);
  for (int i = 1; i <= n; i++) read(p[i]);
  for (int i = 1; i <= n; i++) read(s[i]);
  for (int i = 1; i <= n; i++) read(b[i]);
  for (int i = 1; i <= m; i++) read(inc[i]);
  for (int i = 1; i <= m; i++) read(pref[i]);
}
inline void Discrete() {
  A.push_back(-(1LL << 60));
  B.push_back(-(1LL << 60));
  for (int i = 1; i <= n; i++) {
    que.push_back({p[i], i});
    que.push_back({s[i], i + maxn * 2});
    A.push_back(p[i] + b[i]);
    B.push_back(p[i] - b[i]);
  }
  for (int i = 1; i <= m; i++) {
    que.push_back({inc[i], i + maxn});
    A.push_back(inc[i] + pref[i]);
    B.push_back(inc[i] - pref[i]);
  }
  sort((que.begin()), (que.end())), sort((A.begin()), (A.end())),
      sort((B.begin()), (B.end()));
  A.erase(unique((A.begin()), (A.end())), A.end());
  B.erase(unique((B.begin()), (B.end())), B.end());
}
inline void Sweep() {
  int num = 0;
  for (auto i : que) {
    int x = i.second, y;
    if (x < maxn) {
      num++;
      y = lower_bound((A.begin()), (A.end()), p[x] + b[x]) - A.begin();
      bitA.Modify(y, 1);
      y = lower_bound((B.begin()), (B.end()), p[x] - b[x]) - B.begin();
      bitB.Modify(y, 1);
    } else if (x < maxn * 2) {
      x -= maxn;
      y = lower_bound((A.begin()), (A.end()), inc[x] + pref[x]) - A.begin();
      ans[x] += bitA.Query(y);
      y = lower_bound((B.begin()), (B.end()), inc[x] - pref[x]) - B.begin();
      ans[x] += bitB.Query(y);
      ans[x] -= num;
    } else {
      x -= maxn * 2;
      num--;
      y = lower_bound((A.begin()), (A.end()), p[x] + b[x]) - A.begin();
      bitA.Modify(y, -1);
      y = lower_bound((B.begin()), (B.end()), p[x] - b[x]) - B.begin();
      bitB.Modify(y, -1);
    }
  }
  for (int i = 1; i <= m; i++) printf("%d%c", ans[i], " \n"[i == m]);
}
int main() {
  Read();
  Discrete();
  Sweep();
  return 0;
}