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;
inline long long read() {
  long long f = 1, x = 0;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
const int MAXN = 1e5 + 5;
struct BIT {
  int N, c[MAXN << 1];
  inline int lowbit(int x) { return x & (-x); }
  void add(int p, int x) {
    for (; p <= N; p += lowbit(p)) c[p] += x;
  }
  int ask(int p) {
    int res = 0;
    for (; p > 0; p -= lowbit(p)) res += c[p];
    return res;
  }
} T;
int n, m, p[MAXN], s[MAXN], b[MAXN], inc[MAXN], pref[MAXN], p1[MAXN], p2[MAXN],
    p3[MAXN << 1], ans[MAXN];
void lisanhua(int *p1, int *p2, int *p3) {
  vector<int> v;
  for (int i = 1; i <= n; ++i) v.push_back(p1[i]);
  for (int i = 1; i <= m; ++i) v.push_back(p2[i]);
  sort(v.begin(), v.end());
  v.erase(unique(v.begin(), v.end()), v.end());
  for (int i = 1; i <= n; ++i)
    p3[i] = lower_bound(v.begin(), v.end(), p1[i]) - v.begin() + 1;
  for (int i = 1; i <= m; ++i)
    p3[i + n] = lower_bound(v.begin(), v.end(), p2[i]) - v.begin() + 1;
}
struct node {
  int x, y, z, typ, id;
} q[MAXN << 1], tmp[MAXN << 1];
bool cmp1(node a, node b) {
  return (a.x < b.x) || (a.x == b.x && a.y < b.y) ||
         (a.x == b.x && a.y == b.y && a.z < b.z) ||
         (a.x == b.x && a.y == b.y && a.z == b.z && a.typ > b.typ);
}
bool cmp2(node a, node b) { return a.y <= b.y; }
void cdq(int l, int r) {
  if (l == r) return;
  int mid = (l + r) >> 1;
  cdq(l, mid);
  cdq(mid + 1, r);
  int pa = l, push_back = mid + 1, ct = 0;
  while (pa <= mid && push_back <= r) {
    if (cmp2(q[pa], q[push_back])) {
      if (q[pa].typ == 1) T.add(q[pa].z, 1);
      tmp[++ct] = q[pa++];
    } else {
      if (q[push_back].typ == 0) ans[q[push_back].id] += T.ask(q[push_back].z);
      tmp[++ct] = q[push_back++];
    }
  }
  while (pa <= mid) {
    if (q[pa].typ == 1) T.add(q[pa].z, 1);
    tmp[++ct] = q[pa++];
  }
  while (push_back <= r) {
    if (q[push_back].typ == 0) ans[q[push_back].id] += T.ask(q[push_back].z);
    tmp[++ct] = q[push_back++];
  }
  for (int i = l; i <= mid; ++i) {
    if (q[i].typ == 1) T.add(q[i].z, -1);
  }
  for (int i = 1; i <= ct; ++i) q[l + i - 1] = tmp[i];
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  n = read();
  m = read();
  T.N = n + m;
  for (int i = 1; i <= n; ++i) p[i] = read();
  for (int i = 1; i <= n; ++i) s[i] = read();
  for (int i = 1; i <= n; ++i) b[i] = read();
  for (int i = 1; i <= m; ++i) inc[i] = read();
  for (int i = 1; i <= m; ++i) pref[i] = read();
  for (int i = 1; i <= n; ++i) p1[i] = b[i] + p[i];
  for (int i = 1; i <= m; ++i) p2[i] = inc[i] + pref[i];
  lisanhua(p1, p2, p3);
  for (int i = 1; i <= n + m; ++i) q[i].x = p3[i];
  for (int i = 1; i <= n; ++i) p1[i] = p[i] - b[i];
  for (int i = 1; i <= m; ++i) p2[i] = inc[i] - pref[i];
  lisanhua(p1, p2, p3);
  for (int i = 1; i <= n + m; ++i) q[i].y = p3[i];
  for (int i = 1; i <= n; ++i) p1[i] = -s[i];
  for (int i = 1; i <= m; ++i) p2[i] = -inc[i];
  lisanhua(p1, p2, p3);
  for (int i = 1; i <= n + m; ++i) q[i].z = p3[i];
  for (int i = 1; i <= n; ++i) q[i].typ = 1, q[i].id = i;
  for (int i = 1; i <= m; ++i) q[i + n].id = i;
  sort(q + 1, q + n + m + 1, cmp1);
  cdq(1, n + m);
  for (int i = 1; i <= m; ++i) cout << ans[i] << " ";
  cout << endl;
  return 0;
}