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;
struct node {
  int t, pmb, pab, s, p;
  node() {}
  node(int t, int pmb, int pab, int s, int p)
      : t(t), pmb(pmb), pab(pab), s(s), p(p) {}
};
bool cmp1(node const& a, node const& b) {
  if (a.pmb != b.pmb)
    return a.pmb < b.pmb;
  else
    return a.t < b.t;
}
bool cmp2(node const& a, node const& b) {
  if (a.pab != b.pab)
    return a.pab < b.pab;
  else
    return a.t < b.t;
}
map<int, int> M;
vector<node> v, tmp;
vector<int> dis;
int n, m;
int p[100005], s[100005], b[100005], inc[100005], pref[100005];
int res[100005];
int bits[300005];
void add(int l, int r, int val) {
  for (int i = r; i > 0; i -= (i & -i)) bits[i] += val;
  for (int i = l - 1; i > 0; i -= (i & -i)) bits[i] -= val;
}
int query(int p, int n) {
  int res = 0;
  for (int i = p; i <= n; i += (i & -i)) res += bits[i];
  return res;
}
void solve(int l, int r) {
  int mid = (l + r) / 2;
  tmp.clear();
  for (int i = (l); i <= (mid); i++) {
    if (v[i].t == 0) tmp.push_back(v[i]);
  }
  for (int i = (mid + 1); i <= (r); i++) {
    if (v[i].t == 1) tmp.push_back(v[i]);
  }
  sort(tmp.begin(), tmp.end(), cmp2);
  M.clear();
  dis.clear();
  for (auto i : tmp) {
    if (i.t == 0)
      dis.emplace_back(i.s), dis.emplace_back(i.p);
    else
      dis.emplace_back(i.s);
  }
  int k = 0;
  sort(dis.begin(), dis.end());
  for (auto i : dis) {
    if (!M[i]) M[i] = ++k;
  }
  for (int i = (0); i < (tmp.size()); i++) {
    if (tmp[i].t == 0)
      tmp[i].s = M[tmp[i].s], tmp[i].p = M[tmp[i].p];
    else
      tmp[i].s = M[tmp[i].s];
  }
  for (auto i : tmp) {
    if (i.t == 0 && i.p <= i.s)
      add(i.p, i.s, 1);
    else
      res[i.p] += query(i.s, k);
  }
  for (auto i : tmp) {
    if (i.t == 0 && i.p <= i.s) add(i.p, i.s, -1);
  }
  if (l < mid) solve(l, mid);
  if (mid + 1 < r) solve(mid + 1, r);
}
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", &pref[i]);
  for (int i = (0); i < (n); i++) {
    v.emplace_back(0, p[i] - b[i], p[i] + b[i], s[i], p[i]);
  }
  for (int i = (0); i < (m); i++) {
    v.emplace_back(1, inc[i] - pref[i], inc[i] + pref[i], inc[i], i);
  }
  sort(v.begin(), v.end(), cmp1);
  solve(0, v.size() - 1);
  for (int i = (0); i < (m); i++) printf("%d ", res[i]);
  puts("");
  return 0;
}