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, p[100005], s[100005], b[100005], inc[100005], pre[100005],
    bitl[5 * 100005], bitr[5 * 100005], ans[100005];
struct Query {
  int x, y, add;
  bool l, r;
} e[5 * 100005];
int dem = 0;
vector<int> c1, c2;
int cmp(Query a, Query b) {
  if (a.y != b.y) return a.y < b.y;
  if (a.x != b.x) return a.x < b.x;
  return (a.l | a.r) > (b.l | b.r);
}
void Update(int u, int *bit, int v) {
  for (int i = u; i <= dem; i += i & (-i)) bit[i] += v;
}
int GetBit(int u, int *bit) {
  int ans = 0;
  for (int i = u; i >= 1; i -= i & (-i)) ans = ans + bit[i];
  return ans;
}
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", &pre[i]);
  for (int i = 1; i <= n; i++) {
    e[++dem] = {b[i], p[i], 1, 1, 0};
    e[++dem] = {b[i] - s[i] + p[i] - 1, s[i] + 1, -1, 1, 0};
    e[++dem] = {b[i] + 1, p[i], -1, 0, 1};
    e[++dem] = {b[i] + s[i] - p[i] + 2, s[i] + 1, 1, 0, 1};
    c1.push_back(b[i] + p[i]);
    c1.push_back(b[i] - s[i] + p[i] - 1 + s[i] + 1);
    c2.push_back(b[i] + 1 - p[i]);
    c2.push_back(b[i] + s[i] - p[i] + 2 - s[i] - 1);
  }
  for (int i = 1; i <= m; i++) {
    e[++dem] = {pre[i], inc[i], i, 0, 0};
    c1.push_back(pre[i] + inc[i]);
    c2.push_back(pre[i] - inc[i]);
  }
  sort(c1.begin(), c1.end());
  sort(c2.begin(), c2.end());
  sort(e + 1, e + dem + 1, cmp);
  for (int i = 1; i <= dem; i++) {
    int x = lower_bound(c1.begin(), c1.end(), e[i].x + e[i].y) - c1.begin() + 1;
    int y = lower_bound(c2.begin(), c2.end(), e[i].x - e[i].y) - c2.begin() + 1;
    if (!(e[i].l | e[i].r)) {
      ans[e[i].add] = GetBit(x, bitl) + GetBit(y, bitr);
    } else if (e[i].l) {
      Update(x, bitl, e[i].add);
    } else {
      Update(y, bitr, e[i].add);
    }
  }
  for (int i = 1; i <= m; i++) printf("%d ", ans[i]);
}