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 maxn = 1e5 + 10;
const int maxm = 1e6 + 10;
int p[maxn], s[maxn], b[maxn];
int inc[maxm], pref[maxm];
int ans[maxm];
int t[maxn << 2], e;
struct opt {
  int type, x, y, val;
  opt() {}
  opt(int type, int x, int y, int val) : type(type), x(x), y(y), val(val) {}
};
bool cmp(const opt& a, const opt& b) {
  if (a.x == b.x) {
    if (a.y == b.y) return a.type < b.type;
    return a.y > b.y;
  }
  return a.x < b.x;
}
vector<opt> q;
int bit1[maxm << 2], bit2[maxn << 2];
void update(int* a, int x, int val) {
  while (x) {
    a[x] += val;
    x -= (x & -x);
  }
}
int ask(int* a, int x) {
  int res = 0;
  while (x <= e) {
    res += a[x];
    x += (x & -x);
  }
  return res;
}
int pos(int x) { return lower_bound(t + 1, t + 1 + e, x) - t; }
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", &p[i]);
  for (int i = 1; i <= n; i++) scanf("%d", &s[i]);
  for (int i = 1; i <= n; i++) scanf("%d", &b[i]);
  for (int i = 1; i <= m; i++) scanf("%d", &inc[i]);
  for (int i = 1; i <= m; i++) scanf("%d", &pref[i]);
  for (int i = 1; i <= n; i++) {
    q.push_back(opt(0, p[i], b[i], 1));
    t[++e] = b[i] - p[i];
    q.push_back(opt(1, p[i], b[i] - 1, -1));
    t[++e] = p[i] + b[i] - 1;
    q.push_back(opt(0, s[i] + 1, b[i] + s[i] - p[i] + 1, -1));
    q.push_back(opt(1, s[i] + 1, b[i] - s[i] + p[i] - 2, 1));
  }
  for (int i = 1; i <= m; i++) {
    q.push_back(opt(2, inc[i], pref[i], i));
    t[++e] = inc[i] + pref[i];
    t[++e] = pref[i] - inc[i];
  }
  sort(t + 1, t + 1 + e);
  e = unique(t + 1, t + 1 + e) - (t + 1);
  sort(q.begin(), q.end(), cmp);
  for (int i = 0; i < (int)q.size(); i++) {
    opt u = q[i];
    if (u.type == 0) {
      update(bit1, pos(u.y - u.x), u.val);
    } else if (u.type == 1) {
      update(bit2, pos(u.y + u.x), u.val);
    } else {
      int t1 = ask(bit1, pos(u.y - u.x));
      int t2 = ask(bit2, pos(u.y + u.x));
      ans[u.val] = t1 + t2;
    }
  }
  for (int i = 1; i <= m; i++) printf("%d ", ans[i]);
  return 0;
}