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 max_n = 200005;
int p[max_n], s[max_n], b[max_n], inc[max_n], pref[max_n];
int n, m;
struct Node {
  int x, y, z, f, id;
  bool operator<(Node p) const {
    if (x == p.x) {
      if (y == p.y)
        return z == p.z ? f < p.f : z < p.z;
      else
        return y < p.y;
    } else
      return x < p.x;
  }
} a[max_n];
int a1[max_n], a2[max_n], a3[max_n];
int h[max_n];
void work(int *a1, int *a2, int *a3) {
  for (int i = 1; i <= n; i++) a3[i] = a1[i];
  for (int i = 1; i <= m; i++) a3[i + n] = a2[i];
  for (int i = 1; i <= n + m; i++) h[i] = a3[i];
  sort(h + 1, h + n + m + 1);
  int tot = unique(h + 1, h + n + m + 1) - h - 1;
  for (int i = 1; i <= n + m; i++)
    a3[i] = lower_bound(h + 1, h + tot + 1, a3[i]) - h;
}
int bit[max_n];
void add(int i, int x) {
  for (; i < max_n; i += i & -i) bit[i] += x;
}
int sum(int i) {
  int res = 0;
  for (; i; i -= i & -i) res += bit[i];
  return res;
}
bool cmp(Node p1, Node p2) {
  if (p1.y == p2.y)
    return p1.z == p2.z ? p1.f < p2.f : p1.z < p2.z;
  else
    return p1.y < p2.y;
}
int ans[max_n];
void cdq(int l, int r) {
  if (l == r) return;
  int mid = l + r >> 1;
  cdq(l, mid), cdq(mid + 1, r);
  sort(a + l, a + mid + 1, cmp), sort(a + mid + 1, a + r + 1, cmp);
  int i = l;
  for (int j = mid + 1; j <= r; j++) {
    while (i <= mid && a[i].y <= a[j].y) {
      if (!a[i].f) add(a[i].z, 1);
      i++;
    }
    if (a[j].f == 1) ans[a[j].id] += sum(a[j].z);
  }
  for (i--; i >= l; i--)
    if (!a[i].f) add(a[i].z, -1);
}
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", 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++) a1[i] = b[i] + p[i];
  for (int i = 1; i <= m; i++) a2[i] = inc[i] + pref[i];
  work(a1, a2, a3);
  for (int i = 1; i <= n + m; i++) a[i].x = a3[i];
  for (int i = 1; i <= n; i++) a1[i] = p[i] - b[i];
  for (int i = 1; i <= m; i++) a2[i] = inc[i] - pref[i];
  work(a1, a2, a3);
  for (int i = 1; i <= n + m; i++) a[i].y = a3[i];
  for (int i = 1; i <= n; i++) a1[i] = -s[i];
  for (int i = 1; i <= m; i++) a2[i] = -inc[i];
  work(a1, a2, a3);
  for (int i = 1; i <= n + m; i++) a[i].z = a3[i];
  for (int i = 1; i <= m; i++) a[n + i].f = 1, a[n + i].id = i;
  sort(a + 1, a + n + m + 1);
  cdq(1, n + m);
  for (int i = 1; i <= m; i++) printf("%d ", ans[i]);
  printf("\n");
  return 0;
}