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 = 2e5 + 5;
int num1, num2;
int a[maxn], b[maxn], ans[maxn];
int pri[maxn], s[maxn], be[maxn], inc[maxn], pref[maxn];
struct node {
  int op, x, y, z, id;
} q[maxn * 2];
bool cmp(node q1, node q2) {
  return q1.x == q2.x ? q1.op < q2.op : q1.x < q2.x;
}
struct BIT {
  int c[maxn];
  int lowbit(int x) { return x & (-x); }
  void update(int k, int x) {
    for (int i = k; i <= max(num1, num2); i += lowbit(i)) c[i] += x;
  }
  int query(int k) {
    int res = 0;
    for (int i = k; i; i -= lowbit(i)) res += c[i];
    return res;
  }
} T[2];
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  int cnt = 0;
  for (int i = 1; i <= n; i++) scanf("%d", &pri[i]);
  for (int i = 1; i <= n; i++) scanf("%d", &s[i]);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &be[i]);
    a[++num1] = pri[i] + be[i];
    b[++num2] = pri[i] - be[i];
    q[++cnt] = {0, pri[i], a[num1], b[num2], i};
    q[++cnt] = {1, s[i] + 1, a[num1], b[num2], i};
  }
  for (int i = 1; i <= m; i++) scanf("%d", &inc[i]);
  for (int i = 1; i <= m; i++) {
    scanf("%d", &pref[i]);
    a[++num1] = inc[i] + pref[i];
    b[++num2] = inc[i] - pref[i];
    q[++cnt] = {2, inc[i], a[num1], b[num2], i};
  }
  sort(a + 1, a + num1 + 1);
  num1 = unique(a + 1, a + num1 + 1) - a - 1;
  sort(b + 1, b + num2 + 1);
  num2 = unique(b + 1, b + num2 + 1) - b - 1;
  sort(q + 1, q + cnt + 1, cmp);
  int tmp = 0;
  for (int i = 1; i <= cnt; i++) {
    q[i].y = lower_bound(a + 1, a + num1 + 1, q[i].y) - a;
    q[i].z = lower_bound(b + 1, b + num2 + 1, q[i].z) - b;
    if (q[i].op == 0) {
      T[0].update(q[i].y, 1);
      T[1].update(q[i].z, 1);
      tmp++;
    } else if (q[i].op == 1) {
      T[0].update(q[i].y, -1);
      T[1].update(q[i].z, -1);
      tmp--;
    } else {
      ans[q[i].id] = T[0].query(q[i].y) + T[1].query(q[i].z) - tmp;
    }
  }
  for (int i = 1; i <= m; i++) printf("%d ", ans[i]);
  printf("\n");
  return 0;
}