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>
int n, m, p[100005], s[100005], b[100005], inc[100005], pref[100005];
struct tree {
  int c[100005 << 2];
  inline void upd(int x, int v) {
    for (; x <= 400000; x += x & -x) c[x] += v;
  }
  inline int qry(int x, int ret = 0) {
    for (; x; x -= x & -x) ret += c[x];
    return ret;
  }
} T1, T2;
std::pair<int, int> q[100005 << 2];
int cnt, lcnt, rcnt, L[100005 << 2], R[100005 << 2], ans[100005];
int main() {
  std::cin >> n >> m;
  for (int i = 1, iend = n; i <= iend; ++i) std::cin >> p[i];
  for (int i = 1, iend = n; i <= iend; ++i) std::cin >> s[i];
  for (int i = 1, iend = n; i <= iend; ++i) std::cin >> b[i];
  for (int i = 1, iend = m; i <= iend; ++i) std::cin >> inc[i];
  for (int i = 1, iend = m; i <= iend; ++i) std::cin >> pref[i];
  for (int i = 1, iend = n; i <= iend; ++i) {
    q[++cnt] = {p[i], i};
    q[++cnt] = {s[i], i + (400000 << 1)};
    L[++lcnt] = p[i] - b[i], R[++rcnt] = p[i] + b[i];
  }
  for (int i = 1, iend = m; i <= iend; ++i) {
    q[++cnt] = {inc[i], i + 400000};
    L[++lcnt] = inc[i] - pref[i], R[++rcnt] = inc[i] + pref[i];
  }
  L[++lcnt] = -2e9, R[++rcnt] = -2e9;
  std::sort(q + 1, q + cnt + 1);
  std::sort(L + 1, L + lcnt + 1),
      lcnt = std::unique(L + 1, L + lcnt + 1) - L - 1;
  std::sort(R + 1, R + rcnt + 1),
      rcnt = std::unique(R + 1, R + rcnt + 1) - R - 1;
  int tmp = 0;
  for (int i = 1, iend = cnt; i <= iend; ++i) {
    int ty = q[i].second;
    if (ty < 400000) {
      ++tmp;
      T1.upd(std::lower_bound(L + 1, L + lcnt + 1, p[ty] - b[ty]) - L, 1);
      T2.upd(std::lower_bound(R + 1, R + rcnt + 1, p[ty] + b[ty]) - R, 1);
    } else if (ty < (400000 << 1)) {
      ty -= 400000;
      ans[ty] +=
          T1.qry(std::lower_bound(L + 1, L + lcnt + 1, inc[ty] - pref[ty]) - L);
      ans[ty] +=
          T2.qry(std::lower_bound(R + 1, R + rcnt + 1, inc[ty] + pref[ty]) - R);
      ans[ty] -= tmp;
    } else {
      --tmp, ty -= 400000 << 1;
      T1.upd(std::lower_bound(L + 1, L + lcnt + 1, p[ty] - b[ty]) - L, -1);
      T2.upd(std::lower_bound(R + 1, R + rcnt + 1, p[ty] + b[ty]) - R, -1);
    }
  }
  for (int i = 1, iend = m; i <= iend; ++i)
    std::cout << ans[i] << " \n"[i == m];
  return 0;
}