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;
using LL = long long;
namespace _buff {
const size_t BUFF = 1 << 19;
char buff[BUFF], *begin = buff, *end = buff;
char getc() {
  if (begin == end) {
    begin = buff;
    end = buff + fread(buff, 1, BUFF, stdin);
  }
  return begin == end ? -1 : *begin++;
}
}  // namespace _buff
LL read() {
  using namespace _buff;
  LL ret = 0;
  bool pos = true;
  char c = getc();
  for (; (c < '0' || c > '9') && c != '-'; c = getc()) {
    assert(~c);
  }
  if (c == '-') {
    pos = false;
    c = getc();
  }
  for (; c >= '0' && c <= '9'; c = getc()) {
    ret = (ret << 3) + (ret << 1) + (c ^ 48);
  }
  return pos ? ret : -ret;
}
const size_t N = 1e5 + 5;
struct Fenwick {
  vector<int> vc, fen;
  void add1(int val) { vc.emplace_back(val); }
  void prep() {
    sort(vc.begin(), vc.end());
    vc.erase(unique(vc.begin(), vc.end()), vc.end());
    fen.resize(vc.size() + 1);
  }
  void add2(int p) {
    p = upper_bound(vc.begin(), vc.end(), p) - vc.begin();
    for (; p < fen.size(); p += p & (-p)) {
      fen[p] += 1;
    }
  }
  int query(int p) {
    p = upper_bound(vc.begin(), vc.end(), p) - vc.begin();
    int ret = 0;
    for (; p; p ^= p & (-p)) {
      ret += fen[p];
    }
    return ret;
  }
} fen[2][N];
int n, m, p[N], s[N], b[N], inc[N], pref[N], ans[N];
int indf[N], indp[N];
void add1(int id, int p, int val) {
  for (; p <= n; p += p & (-p)) {
    fen[id][p].add1(val);
  }
}
void prep() {
  for (int i = 1; i <= n; ++i) {
    fen[0][i].prep();
    fen[1][i].prep();
  }
}
void add2(int id, int p, int val) {
  for (; p <= n; p += p & (-p)) {
    fen[id][p].add2(val);
  }
}
int query(int id, int p, int val) {
  int ret = 0;
  for (; p; p ^= p & (-p)) {
    ret += fen[id][p].query(val);
  }
  return ret;
}
int main() {
  n = read();
  m = read();
  generate_n(p, n, read);
  generate_n(s, n, read);
  generate_n(b, n, read);
  generate_n(inc, m, read);
  generate_n(pref, m, read);
  iota(indf, indf + n, 0);
  sort(indf, indf + n, [&](int x, int y) { return s[x] > s[y]; });
  iota(indp, indp + m, 0);
  sort(indp, indp + m, [&](int x, int y) { return inc[x] > inc[y]; });
  vector<int> vcp, vcn;
  for (int i = 0; i < n; ++i) {
    vcp.emplace_back(+b[i]);
    vcn.emplace_back(-b[i]);
  }
  sort(vcp.begin(), vcp.end());
  vcp.erase(unique(vcp.begin(), vcp.end()), vcp.end());
  sort(vcn.begin(), vcn.end());
  vcn.erase(unique(vcn.begin(), vcn.end()), vcn.end());
  for (int i = 0, j = 0; i < m; ++i) {
    for (; j < n && s[indf[j]] >= inc[indp[i]]; ++j) {
      add1(0, upper_bound(vcp.begin(), vcp.end(), +b[indf[j]]) - vcp.begin(),
           p[indf[j]] - b[indf[j]]);
      add1(1, upper_bound(vcn.begin(), vcn.end(), -b[indf[j]]) - vcn.begin(),
           p[indf[j]] + b[indf[j]]);
    }
  }
  prep();
  for (int i = 0, j = 0; i < m; ++i) {
    for (; j < n && s[indf[j]] >= inc[indp[i]]; ++j) {
      add2(0, upper_bound(vcp.begin(), vcp.end(), +b[indf[j]]) - vcp.begin(),
           p[indf[j]] - b[indf[j]]);
      add2(1, upper_bound(vcn.begin(), vcn.end(), -b[indf[j]]) - vcn.begin(),
           p[indf[j]] + b[indf[j]]);
    }
    ans[indp[i]] =
        query(0,
              upper_bound(vcp.begin(), vcp.end(), +pref[indp[i]]) - vcp.begin(),
              inc[indp[i]] - pref[indp[i]]) +
        query(1,
              lower_bound(vcn.begin(), vcn.end(), -pref[indp[i]]) - vcn.begin(),
              inc[indp[i]] + pref[indp[i]]);
  }
  for (int i = 0; i < m; ++i) {
    printf("%d%c", ans[i], "\n "[i < n]);
  }
  return 0;
}