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 long long maxn = 2e5 + 5;
const long long mod = 1e9 + 7;
const long long inf = 1e9 + 1;
long long ans;
long long n, m, p[maxn], s[maxn], b[maxn], inc[maxn], pref[maxn];
vector<long long> v1[maxn * 4], v2[maxn * 4], vec;
void compress() {
  sort(vec.begin(), vec.end());
  vec.resize(unique(vec.begin(), vec.end()) - vec.begin());
  for (long long i = 0; i < n; i++) {
    p[i] = lower_bound(vec.begin(), vec.end(), p[i]) - vec.begin();
    s[i] = lower_bound(vec.begin(), vec.end(), s[i]) - vec.begin();
  }
}
void add(long long l, long long r, long long x1, long long x2, long long v = 1,
         long long s = 0, long long e = (long long)vec.size()) {
  if ((l >= e) || (r <= s)) return;
  if ((l <= s) && (e <= r)) {
    v1[v].push_back(x1);
    v2[v].push_back(x2);
    return;
  }
  long long mid = (e + s) >> 1, lc = v << 1, rc = lc | 1;
  add(l, r, x1, x2, lc, s, mid);
  add(l, r, x1, x2, rc, mid, e);
}
void SRT(long long v = 1, long long s = 0,
         long long e = (long long)vec.size()) {
  sort(v1[v].begin(), v1[v].end());
  sort(v2[v].begin(), v2[v].end());
  if (e - s == 1) return;
  long long mid = (e + s) >> 1, lc = v << 1, rc = lc | 1;
  SRT(lc, s, mid);
  SRT(rc, mid, e);
}
long long get(long long ind, long long j, long long v = 1, long long s = 0,
              long long e = (long long)vec.size()) {
  long long cnt = 0;
  if (v1[v].size()) {
    long long l = -1, r = v1[v].size();
    while (l != r - 1) {
      long long mid = (l + r) / 2;
      if (v1[v][mid] > pref[j] + inc[j])
        r = mid;
      else
        l = mid;
    }
    cnt += (long long)v1[v].size() - r;
    long long r1 = r;
    l = -1, r = v1[v].size();
    while (l != r - 1) {
      long long mid = (l + r) / 2;
      if (v2[v][mid] < pref[j] - inc[j])
        l = mid;
      else
        r = mid;
    }
    cnt += r;
    cnt = (long long)v1[v].size() - cnt;
  }
  if (e - s == 1) return cnt;
  long long mid = (e + s) >> 1, lc = v << 1, rc = lc | 1;
  if (ind < mid) return cnt + get(ind, j, lc, s, mid);
  return cnt + get(ind, j, rc, mid, e);
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n >> m;
  for (long long i = 0; i < n; i++) cin >> p[i];
  for (long long i = 0; i < n; i++) {
    cin >> s[i];
    vec.push_back(s[i]);
    vec.push_back(p[i]);
    vec.push_back(s[i] + 1);
  }
  for (long long i = 0; i < n; i++) cin >> b[i];
  for (long long i = 0; i < m; i++) cin >> inc[i];
  for (long long i = 0; i < m; i++) cin >> pref[i];
  compress();
  for (long long i = 0; i < n; i++) {
    long long x = b[i];
    long long y = vec[p[i]];
    long long l = x + y;
    long long r = x - y;
    add(p[i], s[i] + 1, l, r);
  }
  SRT();
  for (long long i = 0; i < m; i++) {
    if ((inc[i] < vec[0]) || (inc[i] >= vec.back())) {
      cout << 0 << ' ';
      continue;
    }
    long long x =
        (upper_bound(vec.begin(), vec.end(), inc[i]) - vec.begin()) - 1ll;
    cout << get(x, i) << ' ';
  }
}