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;
using ull = unsigned long long;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
const int MN = 4e5 + 3, MM = 2e5 + 3, MOD = 1e9 + 7;
mt19937_64 rnd =
    mt19937_64(chrono::steady_clock::now().time_since_epoch().count());
struct Dish {
  int p, s, b, v1, v2;
};
struct Dude {
  int inc, pref, ind, v1, v2;
};
int ans[MN], bit[MN];
void inc(int i) {
  for (; i < MN; i += i & -i) ++bit[i];
}
void dec(int i) {
  for (; i < MN; i += i & -i) --bit[i];
}
int query(int i) {
  int ret = 0;
  for (; i; i ^= i & -i) ret += bit[i];
  return ret;
}
vector<int> xs;
void getx(int &first) {
  first = lower_bound((xs).begin(), (xs).end(), first) - xs.begin() + 1;
}
void cdq(int l, int r, vector<Dish> &dishes, vector<Dude> &dudes) {
  if (dishes.empty() || dudes.empty()) return;
  vector<Dish> lq, rq;
  vector<Dude> ld, rd;
  int bp = 0, mid = (l + r) / 2;
  for (Dude &d : dudes) {
    while (bp < dishes.size() && dishes[bp].v1 <= d.v1) {
      if (dishes[bp].b <= mid && l != r)
        lq.push_back(dishes[bp++]);
      else {
        rq.push_back(dishes[bp]);
        inc(dishes[bp].p);
        dec(dishes[bp++].s + 1);
      }
    }
    if (d.pref <= mid || l == r) {
      ans[d.ind] += query(d.inc);
      ld.push_back(d);
    } else
      rd.push_back(d);
  }
  for (Dish &d : rq) dec(d.p), inc(d.s + 1);
  if (l < r) {
    cdq(l, mid, lq, ld);
    cdq(mid + 1, r, rq, rd);
  }
}
void cdq2(int l, int r, vector<Dish> &dishes, vector<Dude> &dudes) {
  if (dishes.empty() || dudes.empty()) return;
  vector<Dish> lq, rq;
  vector<Dude> ld, rd;
  int bp = 0, mid = (l + r) / 2;
  for (Dude &d : dudes) {
    while (bp < dishes.size() && dishes[bp].v2 <= d.v2) {
      if (dishes[bp].b <= mid) {
        lq.push_back(dishes[bp]);
        inc(dishes[bp].p);
        dec(dishes[bp++].s + 1);
      } else
        rq.push_back(dishes[bp++]);
    }
    if (d.pref <= mid)
      ld.push_back(d);
    else {
      ans[d.ind] += query(d.inc);
      rd.push_back(d);
    }
  }
  for (Dish &d : lq) dec(d.p), inc(d.s + 1);
  if (l < r) {
    cdq2(l, mid, lq, ld);
    cdq2(mid + 1, r, rq, rd);
  }
}
int main() {
  int n, m, ree = 0;
  scanf("%d %d", &n, &m);
  vector<Dish> dishes(n);
  vector<Dude> dudes(m);
  for (auto &au : dishes) scanf("%d", &au.p), xs.push_back(au.p);
  for (auto &au : dishes) scanf("%d", &au.s), xs.push_back(au.s);
  for (auto &au : dishes)
    scanf("%d", &au.b), au.v1 = au.b + au.p, au.v2 = au.p - au.b;
  for (auto &au : dudes)
    scanf("%d", &au.inc), xs.push_back(au.inc), au.ind = ++ree;
  for (auto &au : dudes)
    scanf("%d", &au.pref), au.v1 = au.inc + au.pref, au.v2 = au.inc - au.pref;
  sort((dishes).begin(), (dishes).end(),
       [&](const Dish &a, const Dish &b) { return a.v1 < b.v1; });
  sort((dudes).begin(), (dudes).end(),
       [&](const Dude &a, const Dude &b) { return a.v1 < b.v1; });
  sort((xs).begin(), (xs).end());
  xs.erase(unique((xs).begin(), (xs).end()), xs.end());
  for (auto &au : dishes) getx(au.p), getx(au.s);
  for (auto &au : dudes) getx(au.inc);
  xs.clear();
  for (auto &au : dishes) xs.push_back(au.b);
  for (auto &au : dudes) xs.push_back(au.pref);
  sort((xs).begin(), (xs).end());
  xs.erase(unique((xs).begin(), (xs).end()), xs.end());
  for (auto &au : dishes) getx(au.b);
  for (auto &au : dudes) getx(au.pref);
  cdq(1, xs.size(), dishes, dudes);
  sort((dishes).begin(), (dishes).end(),
       [&](const Dish &a, const Dish &b) { return a.v2 < b.v2; });
  sort((dudes).begin(), (dudes).end(),
       [&](const Dude &a, const Dude &b) { return a.v2 < b.v2; });
  cdq2(1, xs.size(), dishes, dudes);
  for (int i = 1; i <= m; i++) printf("%d ", ans[i]);
  printf("\n");
  return 0;
}