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;
int n, m, nrE;
struct Dish {
  int p, s, b;
} d[100002];
struct Person {
  int inc, pref;
} v[100002];
struct Event {
  int x, y, z, t;
  bool operator<(const Event &ot) const {
    if (x == ot.x) return z > ot.z;
    return x < ot.x;
  }
} E[100002 * 3];
unordered_map<int, int> Bit[2];
int ans[100002];
void Add(int t, int i, int val) {
  for (; i < 2 * 1000000005 && i > 0; i += (i & (-i))) Bit[t][i] += val;
}
int Sum(int t, int i) {
  int ret = 0;
  for (; i > 0; i -= (i & (-i))) ret += Bit[t][i];
  return ret;
}
void readDish() {
  for (int i = 0; i < n; ++i) scanf("%d", &d[i].p);
  for (int i = 0; i < n; ++i) scanf("%d", &d[i].s);
  for (int i = 0; i < n; ++i) scanf("%d", &d[i].b);
}
void readPers() {
  for (int i = 0; i < m; ++i) scanf("%d", &v[i].inc);
  for (int i = 0; i < m; ++i) scanf("%d", &v[i].pref);
}
void solveEvents() {
  sort(E, E + nrE);
  for (int i = 0; i < nrE; ++i) {
    if (E[i].z <= 0) {
      int idx = -E[i].z;
      ans[idx] = Sum(0, v[idx].inc + v[idx].pref) +
                 Sum(1, v[idx].pref - v[idx].inc + 1000000005);
    } else {
      Add(0, E[i].x + E[i].y, 1 * E[i].t);
      Add(1, E[i].z - E[i].x + 1000000005, -1 * E[i].t);
    }
  }
}
int main() {
  scanf("%d%d", &n, &m);
  readDish();
  readPers();
  for (int i = 0; i < n + m; ++i) {
    if (i < n) {
      E[nrE++] = {d[i].p, d[i].b, d[i].b + 1, 1};
      E[nrE++] = {d[i].s + 1, d[i].b - (d[i].s - d[i].p + 1),
                  d[i].b + (d[i].s - d[i].p + 1) + 1, -1};
    } else
      E[nrE++] = {v[i - n].inc, v[i - n].pref, n - i, 0};
  }
  solveEvents();
  for (int i = 0; i < m; ++i) printf("%d ", ans[i]);
  return 0;
}