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 = 200005;
struct Node {
  int op, a, b, c;
  int id;
} dat[N], td[N];
inline bool cmp_a(const Node &t1, const Node &t2) {
  return t1.a < t2.a || (t1.a == t2.a && t1.op < t2.op);
}
inline bool cmp_b(const Node &t1, const Node &t2) {
  return t1.b < t2.b || (t1.b == t2.b && t1.op < t2.op);
}
int ans[N];
int bit[N];
inline void Add(int i, int x, int n) {
  while (i <= n) {
    bit[i] += x;
    i += i & -i;
  }
}
inline int Getsum(int i) {
  int res = 0;
  while (i > 0) {
    res += bit[i];
    i -= i & -i;
  }
  return res;
}
void dc(int l, int r, int gay) {
  if (l == r) return;
  int mid = (l + r) >> 1;
  dc(l, mid, gay);
  dc(mid + 1, r, gay);
  int cnt = 0;
  for (int i = l; i <= mid; i++) {
    if (dat[i].op == 1) td[++cnt] = dat[i];
  }
  if (!cnt) return;
  int f_ck = cnt;
  for (int i = mid + 1; i <= r; i++) {
    if (dat[i].op == 2) td[++cnt] = dat[i];
  }
  if (cnt == f_ck) return;
  sort(td + 1, td + 1 + cnt, cmp_b);
  for (int i = 1; i <= cnt; i++)
    if (td[i].op == 1)
      Add(td[i].c, 1, gay);
    else
      ans[td[i].id] += Getsum(td[i].c);
  for (int i = 1; i <= cnt; i++)
    if (td[i].op == 1) Add(td[i].c, -1, gay);
}
vector<int> vec;
inline void solve(int n) {
  sort(dat + 1, dat + 1 + n, cmp_a);
  vector<int>().swap(vec);
  for (int i = 1; i <= n; i++) {
    vec.push_back(dat[i].c);
  }
  sort(vec.begin(), vec.end());
  vec.resize(unique(vec.begin(), vec.end()) - vec.begin());
  for (int i = 1; i <= n; i++) {
    dat[i].c = lower_bound(vec.begin(), vec.end(), dat[i].c) - vec.begin() + 1;
  }
  dc(1, n, (int)vec.size());
}
int n, m;
int p[N], s[N], b[N], inc[N], pref[N];
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", &pref[i]);
  for (int i = 1; i <= n; i++) {
    dat[i].op = 1;
    dat[i].a = -b[i];
    dat[i].b = p[i] + b[i];
    dat[i].c = -s[i];
  }
  for (int i = 1; i <= m; i++) {
    dat[i + n].op = 2;
    dat[i + n].id = i;
    dat[i + n].a = -pref[i];
    dat[i + n].b = inc[i] + pref[i];
    dat[i + n].c = -inc[i];
  }
  solve(n + m);
  for (int i = 1; i <= n; i++) {
    dat[i].op = 1;
    dat[i].a = b[i];
    dat[i].b = p[i] - b[i];
    dat[i].c = -s[i];
  }
  for (int i = 1; i <= m; i++) {
    dat[i + n].op = 2;
    dat[i + n].id = i;
    dat[i + n].a = pref[i] - 1;
    dat[i + n].b = inc[i] - pref[i];
    dat[i + n].c = -inc[i];
  }
  solve(n + m);
  for (int i = 1; i <= m; i++) printf("%d%c", ans[i], " \n"[i == m]);
  return 0;
}