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;
struct data {
  int w, x, y, opt, id;
  inline data() { w = x = y = opt = id = 0; }
  inline data(int _w, int _x, int _y, int _opt, int _id)
      : w(_w), x(_x), y(_y), opt(_opt), id(_id) {}
  inline bool operator<(const data &a) const {
    return w < a.w || w == a.w && id < a.id;
  }
};
int p[400005], s[400005], b[400005], inc[400005], pref[400005];
int a[400005];
data q[400005];
int ans[400005];
struct BIT {
  int val[400005], size;
  inline void update(int k, int w) {
    for (; k <= size; k += k & -k) val[k] += w;
  }
  inline int query(int k) {
    int s = 0;
    for (; k; k -= k & -k) s += val[k];
    return s;
  }
} bit1, bit2;
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]);
  int cnt = 0;
  for (int i = 1; i <= n; i++)
    a[++cnt] = p[i] + b[i], a[++cnt] = b[i] - p[i] + 1;
  for (int i = 1; i <= m; i++)
    a[++cnt] = inc[i] + pref[i], a[++cnt] = pref[i] - inc[i];
  sort(a + 1, a + cnt + 1);
  cnt = unique(a + 1, a + cnt + 1) - a - 1;
  int k = 0;
  for (int i = 1; i <= n; i++)
    q[++k] = data(p[i], p[i], b[i], 1, 0),
    q[++k] = data(s[i] + 1, p[i], b[i], -1, 0);
  for (int i = 1; i <= m; i++) q[++k] = data(inc[i], inc[i], pref[i], 0, i);
  sort(q + 1, q + k + 1);
  bit1.size = bit2.size = cnt;
  for (int i = 1; i <= k; i++) {
    if (q[i].id) {
      int t1 = lower_bound(a + 1, a + cnt + 1, q[i].y - q[i].x) - a,
          t2 = lower_bound(a + 1, a + cnt + 1, q[i].x + q[i].y) - a;
      ans[q[i].id] = bit2.query(t2) - bit1.query(t1);
    } else {
      int t1 = lower_bound(a + 1, a + cnt + 1, q[i].y - q[i].x + 1) - a,
          t2 = lower_bound(a + 1, a + cnt + 1, q[i].x + q[i].y) - a;
      bit1.update(t1, q[i].opt);
      bit2.update(t2, q[i].opt);
    }
  }
  for (int i = 1; i <= m; i++) printf("%d ", ans[i]);
  return 0;
}