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 = 100010, M = 100010;
int n, m, p[N], s[N], b[N], inc[M], pref[M];
int lsx[2 * N + M], lsy[2 * N], endx, endy;
inline int lowbit(int x) { return x & (-x); }
class fwk {
 public:
  int dat[2 * N];
  void modify(int p, int v) {
    for (; p <= endy; p += lowbit(p)) dat[p] += v;
  }
  int query(int p) {
    int ret = 0;
    for (; p; p -= lowbit(p)) ret += dat[p];
    return ret;
  }
} T1, T2;
vector<int> add[2 * N + M], del[2 * N + M], q[2 * N + M];
int ans[M], cnt = 0;
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", &p[i]), lsx[++endx] = p[i];
  for (int i = 1; i <= n; i++) scanf("%d", &s[i]), lsx[++endx] = s[i];
  for (int i = 1; i <= n; i++)
    scanf("%d", &b[i]), lsy[++endy] = p[i] + b[i], lsy[++endy] = p[i] - b[i];
  for (int i = 1; i <= m; i++) scanf("%d", &inc[i]), lsx[++endx] = inc[i];
  for (int i = 1; i <= m; i++) scanf("%d", &pref[i]);
  sort(lsx + 1, lsx + endx + 1);
  sort(lsy + 1, lsy + endy + 1);
  endx = unique(lsx + 1, lsx + endx + 1) - lsx - 1;
  endy = unique(lsy + 1, lsy + endy + 1) - lsy - 1;
  for (int i = 1; i <= n; i++) {
    add[lower_bound(lsx + 1, lsx + endx + 1, p[i]) - lsx].push_back(i);
    del[lower_bound(lsx + 1, lsx + endx + 1, s[i]) - lsx + 1].push_back(i);
  }
  for (int i = 1; i <= m; i++)
    q[lower_bound(lsx + 1, lsx + endx + 1, inc[i]) - lsx].push_back(i);
  for (int i = 1; i <= endx; i++) {
    for (auto &x : add[i]) {
      cnt++;
      T1.modify(lower_bound(lsy + 1, lsy + endy + 1, p[x] + b[x]) - lsy, 1);
      T2.modify(lower_bound(lsy + 1, lsy + endy + 1, p[x] - b[x]) - lsy, 1);
    }
    for (auto &x : del[i]) {
      cnt--;
      T1.modify(lower_bound(lsy + 1, lsy + endy + 1, p[x] + b[x]) - lsy, -1);
      T2.modify(lower_bound(lsy + 1, lsy + endy + 1, p[x] - b[x]) - lsy, -1);
    }
    for (auto &x : q[i]) {
      int val1 = T1.query(
              upper_bound(lsy + 1, lsy + endy + 1, inc[x] + pref[x]) - lsy - 1),
          val2 = T2.query(
              upper_bound(lsy + 1, lsy + endy + 1, inc[x] - pref[x]) - lsy - 1);
      ans[x] = val1 + val2 - cnt;
    }
  }
  for (int i = 1; i <= m; i++) printf("%d ", ans[i]);
  return 0;
}