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, acnt, bcnt, pa, pb, tot;
struct hp {
  int x, y, z, cnt, id;
} f[222222], a[222222], b[222222];
int C[222222], cnt[222222], ch[222222], p1[222222], p2[222222], p3[222222];
int p[222222], s[222222], d[222222], inc[222222], pref[222222];
int cmp1(hp a, hp b) {
  return a.x < b.x || (a.x == b.x && a.y < b.y) ||
         (a.x == b.x && a.y == b.y && a.z < b.z) ||
         (a.x == b.x && a.y == b.y && a.z == b.z && a.cnt > b.cnt);
}
int cmp2(hp a, hp b) { return a.y < b.y; }
void add(int loc, int val) {
  if (!loc) return;
  for (int i = loc; i <= n + m; i += i & (-i)) C[i] += val;
}
int query(int loc) {
  int ans = 0;
  for (int i = loc; i >= 1; i -= i & (-i)) ans += C[i];
  return ans;
}
void cdq(int l, int r) {
  if (l >= r) return;
  int mid = (l + r) >> 1;
  cdq(l, mid);
  acnt = 0;
  for (int i = l; i <= mid; ++i) a[++acnt] = f[i];
  sort(a + 1, a + acnt + 1, cmp2);
  bcnt = 0;
  for (int i = mid + 1; i <= r; ++i) b[++bcnt] = f[i];
  sort(b + 1, b + bcnt + 1, cmp2);
  pa = pb = 1;
  tot = 0;
  while (pb <= bcnt) {
    while (pa <= acnt && a[pa].y <= b[pb].y) {
      if (a[pa].cnt) {
        add(a[pa].z, a[pa].cnt);
        ch[++tot] = a[pa].z;
      }
      ++pa;
    }
    if (b[pb].cnt == 0) cnt[b[pb].id] += query(b[pb].z);
    ++pb;
  }
  for (int i = 1; i <= tot; ++i) add(ch[i], -1);
  cdq(mid + 1, r);
}
int main() {
  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", &d[i]);
  for (int i = 1; i <= m; i++) scanf("%d", &inc[i]);
  for (int i = 1; i <= m; i++) scanf("%d", &pref[i]);
  auto discretize = [&](int *p, int *q, int *x) {
    vector<int> number;
    for (int i = 1; i <= n; i++) number.push_back(p[i]);
    for (int i = 1; i <= m; i++) number.push_back(q[i]);
    sort(number.begin(), number.end());
    number.erase(unique(number.begin(), number.end()), number.end());
    for (int i = 1; i <= n; i++)
      x[i] =
          lower_bound(number.begin(), number.end(), p[i]) - number.begin() + 1;
    for (int i = 1; i <= m; i++)
      x[i + n] =
          lower_bound(number.begin(), number.end(), q[i]) - number.begin() + 1;
  };
  for (int i = 1; i <= n; i++) p1[i] = p[i] + d[i];
  for (int i = 1; i <= m; i++) p2[i] = inc[i] + pref[i];
  discretize(p1, p2, p3);
  for (int i = 1; i <= n + m; i++) f[i].x = p3[i];
  for (int i = 1; i <= n; i++) p1[i] = p[i] - d[i];
  for (int i = 1; i <= m; i++) p2[i] = inc[i] - pref[i];
  discretize(p1, p2, p3);
  for (int i = 1; i <= n + m; i++) f[i].y = p3[i];
  for (int i = 1; i <= n; i++) p1[i] = -s[i];
  for (int i = 1; i <= m; i++) p2[i] = -inc[i];
  discretize(p1, p2, p3);
  for (int i = 1; i <= n + m; i++) f[i].z = p3[i];
  for (int i = 1; i <= n; ++i) {
    f[i].id = i;
    f[i].cnt = 1;
  }
  for (int i = 1; i <= m; ++i) f[i + n].id = i;
  sort(f + 1, f + n + m + 1, cmp1);
  cdq(1, n + m);
  for (int i = 1; i <= m; i++) printf("%d ", cnt[i]);
}