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 = 200005;
int n, m;
int p[MAXN], s[MAXN], b[MAXN], inc[MAXN], pref[MAXN];
int tmp[MAXN], ans[MAXN];
struct Data {
  int ty, x, y, z, id;
  Data() {}
  Data(int ty, int x, int y, int z, int id)
      : ty(ty), x(x), y(y), z(z), id(id) {}
} a[MAXN], d[MAXN];
inline bool cmpx(Data a, Data b) {
  return a.x == b.x ? a.ty > b.ty : a.x < b.x;
}
inline bool cmpy(Data a, Data b) {
  return a.y == b.y ? a.ty > b.ty : a.y < b.y;
}
int c[MAXN];
void add(int x, int d) {
  for (int i = x; i <= n + m; i += (i & -i)) c[i] += d;
}
int query(int x) {
  int res = 0;
  for (int i = x; i; i -= (i & -i)) res += c[i];
  return res;
}
void solve(int l, int r) {
  if (l == r) return;
  int mid = (l + r) >> 1;
  solve(l, mid);
  solve(mid + 1, r);
  for (int i = l; i <= r; i++) {
    d[i] = a[i];
    d[i].x = i;
  }
  sort(d + l, d + r + 1, cmpy);
  for (int i = l; i <= r; i++)
    if (d[i].x <= mid && !d[i].ty)
      add(d[i].z, 1);
    else if (d[i].x > mid && d[i].ty)
      ans[d[i].id] += query(d[i].z - 1);
  for (int i = l; i <= r; i++)
    if (d[i].x <= mid && !d[i].ty) add(d[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++) a[i] = Data(0, -b[i], -s[i], b[i] + p[i], i);
  for (int j = 1; j <= m; j++)
    a[j + n] = Data(1, -pref[j] + 1, -inc[j] + 1, inc[j] + pref[j] + 1, j + n);
  for (int i = 1; i <= n + m; i++) tmp[i] = a[i].x;
  sort(tmp + 1, tmp + n + m + 1);
  for (int i = 1; i <= n + m; i++)
    a[i].x = lower_bound(tmp + 1, tmp + n + m + 1, a[i].x) - tmp;
  for (int i = 1; i <= n + m; i++) tmp[i] = a[i].y;
  sort(tmp + 1, tmp + n + m + 1);
  for (int i = 1; i <= n + m; i++)
    a[i].y = lower_bound(tmp + 1, tmp + n + m + 1, a[i].y) - tmp;
  for (int i = 1; i <= n + m; i++) tmp[i] = a[i].z;
  sort(tmp + 1, tmp + n + m + 1);
  for (int i = 1; i <= n + m; i++)
    a[i].z = lower_bound(tmp + 1, tmp + n + m + 1, a[i].z) - tmp;
  sort(a + 1, a + n + m + 1, cmpx);
  solve(1, n + m);
  for (int i = 1; i <= n; i++) a[i] = Data(0, b[i], -s[i], p[i] - b[i], i);
  for (int j = 1; j <= m; j++)
    a[j + n] = Data(1, pref[j], -inc[j] + 1, inc[j] - pref[j] + 1, j + n);
  for (int i = 1; i <= n + m; i++) tmp[i] = a[i].x;
  sort(tmp + 1, tmp + n + m + 1);
  for (int i = 1; i <= n + m; i++)
    a[i].x = lower_bound(tmp + 1, tmp + n + m + 1, a[i].x) - tmp;
  for (int i = 1; i <= n + m; i++) tmp[i] = a[i].y;
  sort(tmp + 1, tmp + n + m + 1);
  for (int i = 1; i <= n + m; i++)
    a[i].y = lower_bound(tmp + 1, tmp + n + m + 1, a[i].y) - tmp;
  for (int i = 1; i <= n + m; i++) tmp[i] = a[i].z;
  sort(tmp + 1, tmp + n + m + 1);
  for (int i = 1; i <= n + m; i++)
    a[i].z = lower_bound(tmp + 1, tmp + n + m + 1, a[i].z) - tmp;
  sort(a + 1, a + n + m + 1, cmpx);
  solve(1, n + m);
  for (int j = n + 1; j <= n + m; j++) printf("%d ", ans[j]);
  printf("\n");
  return 0;
}