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 N = 2e5 + 5;
int n, m, opsize, lim;
struct OP {
  int tim, d, tag, pos;
} op[N << 2];
inline bool oprule(OP x, OP y) {
  return (x.tim == y.tim) ? (x.tag < y.tag) : (x.tim < y.tim);
}
vector<int> lsh;
int p[N], s[N], b[N], c[N], f[N], ans[N];
int id(long long x) {
  return lower_bound(lsh.begin(), lsh.end(), x) - lsh.begin() + 1;
}
struct BIT {
  int w[N << 2];
  void mdf(int x, int d) {
    x = id(x);
    while (x <= lim) w[x] += d, x += (x & -x);
  }
  int qry(int x) {
    x = id(x);
    int res = 0;
    while (x) res += w[x], x -= (x & -x);
    return res;
  }
} bit[2];
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]);
    op[++opsize] = (OP){p[i], 1, 1, i};
    op[++opsize] = (OP){s[i] + 1, -1, 1, i};
  }
  for (int i = 1; i <= m; ++i) scanf("%d", &c[i]);
  for (int i = 1; i <= m; ++i) {
    scanf("%d", &f[i]);
    op[++opsize] = (OP){c[i], 0, 2, i};
    lsh.push_back(c[i] + f[i]), lsh.push_back(c[i] - f[i]);
  }
  lsh.push_back(2e9), lsh.push_back(-2e9);
  sort(op + 1, op + opsize + 1, oprule);
  sort(lsh.begin(), lsh.end());
  unique(lsh.begin(), lsh.end()), lim = lsh.end() - lsh.begin() + 5;
  for (int i = 1, cnt = 0, pp; i <= opsize; ++i) {
    if (op[i].d)
      cnt += op[i].d, pp = op[i].pos, bit[0].mdf(p[pp] + b[pp], op[i].d),
                      bit[1].mdf(p[pp] - b[pp], op[i].d);
    else
      pp = op[i].pos,
      ans[pp] = bit[0].qry(c[pp] + f[pp]) + bit[1].qry(c[pp] - f[pp]) - cnt;
  }
  for (int i = 1; i <= m; ++i) printf("%d ", ans[i]);
  return 0;
}