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 INF = 0x3f3f3f3f;
int n, m, tot, ans[110000];
map<int, int> t;
struct dish {
  int p, s, b;
} d[110000], u[110000], v[110000];
struct people {
  int inc, pref, id;
  bool operator<(const people &rhs) const { return inc < rhs.inc; }
} a[110000];
int cmp1(const dish &a, const dish &b) { return a.p < b.p; }
int cmp2(const dish &a, const dish &b) { return a.s < b.s; }
struct BIT {
  int n, tot, c[110000];
  map<int, int> t;
  void init() {
    for (auto &e : t) e.second = ++tot;
  }
  void insert(int x) { t[x] = 0; }
  int lowbit(int x) { return x & -x; }
  int sum(int x) {
    int ret = 0;
    while (x) {
      ret += c[x];
      x -= lowbit(x);
    }
    return ret;
  }
  void add(int x, int k) {
    x = t[x];
    while (x <= tot) {
      c[x] += k;
      x += lowbit(x);
    }
  }
  int second(int x) {
    auto it = t.upper_bound(x);
    if (it == t.begin()) return 0;
    it--;
    return sum(it->second);
  }
} B1, B2;
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", &d[i].p);
  for (int i = 1; i <= n; i++) scanf("%d", &d[i].s);
  for (int i = 1; i <= n; i++) scanf("%d", &d[i].b);
  for (int i = 1; i <= m; i++) {
    scanf("%d", &a[i].inc);
    a[i].id = i;
  }
  for (int i = 1; i <= m; i++) scanf("%d", &a[i].pref);
  sort(a + 1, a + m + 1);
  for (int i = 1; i <= n; i++) {
    u[i] = v[i] = d[i];
    B1.insert(d[i].b - d[i].p);
    B2.insert(d[i].b + d[i].p);
  }
  B1.init();
  B2.init();
  sort(u + 1, u + n + 1, cmp1);
  sort(v + 1, v + n + 1, cmp2);
  int t1 = 1, t2 = 1;
  for (int i = 1; i <= m; i++) {
    while (t1 <= n && u[t1].p <= a[i].inc) {
      B1.add(u[t1].b - u[t1].p, 1);
      B2.add(u[t1].b + u[t1].p, 1);
      t1++;
    }
    while (t2 <= n && v[t2].s < a[i].inc) {
      B1.add(v[t2].b - v[t2].p, -1);
      B2.add(v[t2].b + v[t2].p, -1);
      t2++;
    }
    ans[a[i].id] = B2.second(a[i].pref + a[i].inc);
    ans[a[i].id] -= B1.second(a[i].pref - a[i].inc - 1);
  }
  for (int i = 1; i <= m; i++) printf("%d ", ans[i]);
  return 0;
}