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;
int n, m, i, j, p[100005], s[100005], b[100005], inc[100005], pref[100005],
    ans[100005], ss[200005], sizy, sizz;
struct ii {
  char op;
  int x, y, z, val, id;
};
vector<ii> v;
vector<int> all;
bool cmp(ii x, ii y) {
  if (x.x == y.x) return x.op < y.op;
  return x.x < y.x;
}
void add(int x, int y) {
  int i;
  for (i = x; i <= sizz; i += (i & -i)) ss[i] += y;
}
int query(int x) {
  int i, sum = 0;
  for (i = x; i; i -= (i & -i)) sum += ss[i];
  return sum;
}
void solve(vector<ii> v, int l, int r) {
  if (l == r) return;
  if (v.empty()) return;
  int mid = (l + r) / 2;
  vector<ii> vl, vr;
  for (__typeof((v).begin()) it = (v).begin(); it != (v).end(); it++) {
    if (it->y <= mid) {
      vl.push_back(*it);
      if (it->op == 0) {
        add(it->z, it->val);
      }
    } else {
      vr.push_back(*it);
      if (it->op == 1) {
        ans[it->id] += it->val * query(it->z);
      }
    }
  }
  for (__typeof((v).begin()) it = (v).begin(); it != (v).end(); it++) {
    if (it->y <= mid) {
      if (it->op == 0) {
        add(it->z, -it->val);
      }
    }
  }
  solve(vl, l, mid);
  solve(vr, mid + 1, r);
}
int main() {
  scanf("%d%d", &n, &m);
  for ((i) = 1; (i) <= (n); ++(i)) scanf("%d", &p[i]);
  for ((i) = 1; (i) <= (n); ++(i)) scanf("%d", &s[i]);
  for ((i) = 1; (i) <= (n); ++(i)) scanf("%d", &b[i]);
  for ((i) = 1; (i) <= (m); ++(i)) scanf("%d", &inc[i]);
  for ((i) = 1; (i) <= (m); ++(i)) scanf("%d", &pref[i]);
  for ((i) = 1; (i) <= (n); ++(i)) {
    v.push_back((ii){0, s[i] + 1, b[i] + p[i], p[i] - b[i], -1, 0});
    v.push_back((ii){0, p[i], b[i] + p[i], p[i] - b[i], 1, 0});
  }
  for ((i) = 1; (i) <= (m); ++(i)) {
    v.push_back((ii){1, inc[i], inc[i] + pref[i] + 1, inc[i] - pref[i], 1, i});
  }
  sort(v.begin(), v.end(), cmp);
  for ((i) = 0; (i) < (v.size()); ++(i)) all.push_back(v[i].y);
  sort(all.begin(), all.end());
  all.resize(unique(all.begin(), all.end()) - all.begin());
  for ((i) = 0; (i) < (v.size()); ++(i))
    v[i].y = upper_bound(all.begin(), all.end(), v[i].y) - all.begin();
  sizy = all.size();
  all.clear();
  for ((i) = 0; (i) < (v.size()); ++(i)) all.push_back(v[i].z);
  sort(all.begin(), all.end());
  all.resize(unique(all.begin(), all.end()) - all.begin());
  for ((i) = 0; (i) < (v.size()); ++(i))
    v[i].z = upper_bound(all.begin(), all.end(), v[i].z) - all.begin();
  sizz = all.size();
  solve(v, 1, sizy);
  for ((i) = 1; (i) <= (m); ++(i)) {
    printf("%d ", ans[i]);
  }
  return 0;
}