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 MAXN = 2e5 + 10;
struct Splay {
  int tot = 0, root = 0;
  struct Node {
    int son[2], pre, sz, v, tsz;
    void init(int vv, int ssz) {
      v = vv;
      sz = tsz = ssz;
      son[0] = son[1] = pre = 0;
    }
  } tr[MAXN];
  void push_up(int x) {
    tr[x].tsz = tr[x].sz + tr[tr[x].son[0]].tsz + tr[tr[x].son[1]].tsz;
  }
  void rotate(int x) {
    int y = tr[x].pre;
    int z = tr[y].pre;
    int k = x == tr[y].son[1];
    tr[x].pre = z;
    tr[z].son[y == tr[z].son[1]] = x;
    tr[y].son[k] = tr[x].son[k ^ 1];
    tr[tr[y].son[k]].pre = y;
    tr[x].son[k ^ 1] = y;
    tr[y].pre = x;
    push_up(y);
  }
  void splay(int x, int goal) {
    while (tr[x].pre != goal) {
      int y = tr[x].pre;
      int z = tr[y].pre;
      if (z != goal) {
        if ((tr[y].son[1] == x) ^ (tr[z].son[1] == y)) {
          rotate(x);
        } else {
          rotate(y);
        }
      }
      rotate(x);
    }
    if (!goal) root = x;
    push_up(x);
  }
  void insert_point(int v, int z) {
    int p = root;
    int ff = 0;
    while (p && tr[p].v != v) {
      ff = p;
      p = tr[p].son[tr[p].v < v];
    }
    if (p) {
      tr[p].sz += z;
    } else {
      p = ++tot;
      if (ff) tr[ff].son[tr[ff].v < v] = p;
      tr[p].init(v, z);
      tr[p].pre = ff;
    }
    splay(p, 0);
  }
  int find_cnt(int v) {
    insert_point(v, 1);
    int res = tr[tr[root].son[1]].tsz;
    insert_point(v, -1);
    return res;
  }
} S[2];
int tot;
int p[MAXN], s[MAXN], b[MAXN], inc[MAXN], pref[MAXN], ans[MAXN];
struct node {
  int v, id, op;
  bool operator<(const node &rhs) const {
    if (v != rhs.v)
      return v < rhs.v;
    else
      return op < rhs.op;
  }
} no[MAXN * 3];
void add(int v, int id, int op) {
  ++tot;
  no[tot] = node{v, id, op};
}
int main() {
  int n, m;
  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]);
    add(p[i], i, 1);
    add(s[i], i, 3);
  }
  for (int i = 1; i <= m; i++) scanf("%d", &inc[i]);
  for (int i = 1; i <= m; i++) {
    scanf("%d", &pref[i]);
    add(inc[i], i, 2);
  }
  sort(no + 1, no + 1 + tot);
  int sum = 0;
  for (int i = 1; i <= tot; i++) {
    int id = no[i].id;
    int op = no[i].op;
    if (op == 1) {
      sum++;
      S[0].insert_point(p[id] + b[id], 1);
      S[1].insert_point(p[id] - b[id], 1);
    } else if (op == 2) {
      ans[id] = sum - S[0].find_cnt(inc[id] + pref[id]) -
                S[1].find_cnt(inc[id] - pref[id]);
    } else {
      sum--;
      S[0].insert_point(p[id] + b[id], -1);
      S[1].insert_point(p[id] - b[id], -1);
    }
  }
  for (int i = 1; i <= m; i++) printf("%d ", ans[i]);
  return 0;
}