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 MOD = (int)1e9 + 7;
const int FFTMOD = 119 << 23 | 1;
const int INF = (int)1e9 + 23111992;
const long long LINF = (long long)1e18 + 23111992;
const long double PI = acos((long double)-1);
const long double EPS = 1e-9;
inline long long gcd(long long a, long long b) {
  long long r;
  while (b) {
    r = a % b;
    a = b;
    b = r;
  }
  return a;
}
inline long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
inline long long fpow(long long n, long long k, int p = MOD) {
  long long r = 1;
  for (; k; k >>= 1) {
    if (k & 1) r = r * n % p;
    n = n * n % p;
  }
  return r;
}
template <class T>
inline int chkmin(T& a, const T& val) {
  return val < a ? a = val, 1 : 0;
}
template <class T>
inline int chkmax(T& a, const T& val) {
  return a < val ? a = val, 1 : 0;
}
inline unsigned long long isqrt(unsigned long long k) {
  unsigned long long r = sqrt(k) + 1;
  while (r * r > k) r--;
  return r;
}
inline long long icbrt(long long k) {
  long long r = cbrt(k) + 1;
  while (r * r * r > k) r--;
  return r;
}
inline void addmod(int& a, int val, int p = MOD) {
  if ((a = (a + val)) >= p) a -= p;
}
inline void submod(int& a, int val, int p = MOD) {
  if ((a = (a - val)) < 0) a += p;
}
inline int mult(int a, int b, int p = MOD) { return (long long)a * b % p; }
inline int inv(int a, int p = MOD) { return fpow(a, p - 2, p); }
inline int sign(long double x) { return x < -EPS ? -1 : x > +EPS; }
inline int sign(long double x, long double y) { return sign(x - y); }
mt19937 mt(chrono::high_resolution_clock::now().time_since_epoch().count());
inline int myrand() { return abs((int)mt()); }
const int MN = 4e5 + 5;
vector<int> nodes[MN];
vector<int> f[MN];
void fakeUpdate(int u, int v) {
  for (int x = u; x < MN; x += x & -x) nodes[x].push_back(v);
}
void fakeGet(int u, int v) {
  for (int x = u; x > 0; x -= x & -x) nodes[x].push_back(v);
}
void push() {
  for (int x = (0); x < (MN); ++x) {
    sort((nodes[x]).begin(), (nodes[x]).end()),
        (nodes[x]).erase(unique((nodes[x]).begin(), (nodes[x]).end()),
                         (nodes[x]).end());
    f[x].resize(nodes[x].size() + 1);
  }
}
void update(int u, int v) {
  for (int x = u; x < MN; x += x & -x)
    for (int y = lower_bound(nodes[x].begin(), nodes[x].end(), v) -
                 nodes[x].begin() + 1;
         y <= nodes[x].size(); y += y & -y)
      f[x][y]++;
}
int get(int u, int v) {
  int res = 0;
  for (int x = u; x > 0; x -= x & -x)
    for (int y = upper_bound(nodes[x].begin(), nodes[x].end(), v) -
                 nodes[x].begin();
         y > 0; y -= y & -y)
      res += f[x][y];
  return res;
}
void chemthan() {
  int n, m;
  cin >> n >> m;
  vector<int> p(n), s(n), b(n), inc(m), pref(m);
  for (int i = (0); i < (n); ++i) cin >> p[i];
  for (int i = (0); i < (n); ++i) cin >> s[i];
  for (int i = (0); i < (n); ++i) cin >> b[i];
  for (int i = (0); i < (m); ++i) cin >> inc[i];
  for (int i = (0); i < (m); ++i) cin >> pref[i];
  vector<int> dc;
  for (int i = (0); i < (n); ++i)
    dc.push_back(p[i] + b[i]), dc.push_back(p[i] - b[i]);
  for (int i = (0); i < (m); ++i)
    dc.push_back(inc[i] + pref[i]), dc.push_back(inc[i] - pref[i]);
  sort((dc).begin(), (dc).end()),
      (dc).erase(unique((dc).begin(), (dc).end()), (dc).end());
  vector<int> vn, vm;
  for (int i = (0); i < (n); ++i) vn.push_back(i);
  for (int i = (0); i < (m); ++i) vm.push_back(i);
  sort((vn).begin(), (vn).end(), [&](int i, int j) { return s[i] > s[j]; });
  sort((vm).begin(), (vm).end(), [&](int i, int j) { return inc[i] > inc[j]; });
  int ptr = 0;
  for (int e : vm) {
    while (ptr < n && inc[e] <= s[vn[ptr]]) {
      int t = vn[ptr];
      int u = lower_bound((dc).begin(), (dc).end(), p[t] + b[t]) - dc.begin();
      int v = lower_bound((dc).begin(), (dc).end(), p[t] - b[t]) - dc.begin();
      fakeUpdate(u, v);
      ptr++;
    }
    int u =
        lower_bound((dc).begin(), (dc).end(), inc[e] + pref[e]) - dc.begin();
    int v =
        lower_bound((dc).begin(), (dc).end(), inc[e] - pref[e]) - dc.begin();
    fakeGet(u, v);
  }
  push();
  ptr = 0;
  vector<int> res(m);
  for (int e : vm) {
    while (ptr < n && inc[e] <= s[vn[ptr]]) {
      int t = vn[ptr];
      int u = lower_bound((dc).begin(), (dc).end(), p[t] + b[t]) - dc.begin();
      int v = lower_bound((dc).begin(), (dc).end(), p[t] - b[t]) - dc.begin();
      update(u, v);
      ptr++;
    }
    int u =
        lower_bound((dc).begin(), (dc).end(), inc[e] + pref[e]) - dc.begin();
    int v =
        lower_bound((dc).begin(), (dc).end(), inc[e] - pref[e]) - dc.begin();
    res[e] = get(u, v);
  }
  for (int i = (0); i < (m); ++i) {
    cout << res[i] << " \n"[i == m - 1];
  }
}
int main(int argc, char* argv[]) {
  ios_base::sync_with_stdio(0), cin.tie(0);
  if (argc > 1) {
    assert(freopen(argv[1], "r", stdin));
  }
  if (argc > 2) {
    assert(freopen(argv[2], "wb", stdout));
  }
  chemthan();
  cerr << "\nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "ms\n";
  return 0;
}