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 N = 100005;
struct node {
  int L, R, p;
} a[N], b[N];
int n, Q, q[N], ans[N];
vector<int> vC[N * 4], vQ[N * 4];
pair<int, int> qq[N * 2];
void insertQ(int k, int l, int r, int p, int id) {
  vQ[k].push_back(id);
  if (l == r) return;
  int mid = (l + r) / 2;
  if (p <= mid)
    insertQ(k * 2, l, mid, p, id);
  else
    insertQ(k * 2 + 1, mid + 1, r, p, id);
}
void insertC(int k, int l, int r, int id) {
  if (a[id].R < q[l] || a[id].L > q[r]) return;
  if (a[id].L <= q[l] && a[id].R >= q[r]) return vC[k].push_back(id), void(0);
  int mid = (l + r) / 2;
  insertC(k * 2, l, mid, id);
  insertC(k * 2 + 1, mid + 1, r, id);
}
void solve(int k, int l, int r) {
  int top = 0, S = 0;
  for (auto i : vQ[k]) qq[++top] = pair<int, int>(b[i].L + b[i].p, i);
  for (auto i : vC[k]) qq[++top] = pair<int, int>(a[i].L + a[i].p, 0);
  sort(qq + 1, qq + top + 1);
  for (int i = (int)(1); i <= (int)(top); i++)
    if (qq[i].second == 0)
      ++S;
    else
      ans[qq[i].second] += S;
  top = 0;
  S = 0;
  for (auto i : vQ[k]) qq[++top] = pair<int, int>(b[i].L - b[i].p, i);
  for (auto i : vC[k]) qq[++top] = pair<int, int>(a[i].L - a[i].p, 0);
  sort(qq + 1, qq + top + 1);
  for (int i = (int)(top); i >= (int)(1); i--)
    if (qq[i].second == 0)
      ++S;
    else
      ans[qq[i].second] -= S;
  if (l == r) return;
  int mid = (l + r) / 2;
  solve(k * 2, l, mid);
  solve(k * 2 + 1, mid + 1, r);
}
int main() {
  scanf("%d%d", &n, &Q);
  for (int i = (int)(1); i <= (int)(n); i++) scanf("%d", &a[i].L);
  for (int i = (int)(1); i <= (int)(n); i++) scanf("%d", &a[i].R);
  for (int i = (int)(1); i <= (int)(n); i++) scanf("%d", &a[i].p);
  for (int i = (int)(1); i <= (int)(Q); i++) scanf("%d", &b[i].L);
  for (int i = (int)(1); i <= (int)(Q); i++) scanf("%d", &b[i].p);
  for (int i = (int)(1); i <= (int)(Q); i++) q[++*q] = b[i].L;
  sort(q + 1, q + *q + 1);
  *q = unique(q + 1, q + *q + 1) - q - 1;
  for (int i = (int)(1); i <= (int)(Q); i++) {
    int p = lower_bound(q + 1, q + *q + 1, b[i].L) - q;
    insertQ(1, 1, *q, p, i);
  }
  for (int i = (int)(1); i <= (int)(n); i++) insertC(1, 1, *q, i);
  solve(1, 1, *q);
  for (int i = (int)(1); i <= (int)(Q); i++) printf("%d ", ans[i]);
}