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 db = double;
using vec = vector<int>;
using pii = pair<int, int>;
int read() {
  int x = 0;
  bool f = 0;
  char ch = getchar();
  while (!isdigit(ch)) f |= ch == '-', ch = getchar();
  while (isdigit(ch)) x = (x << 1) + (x << 3) + (ch ^ 48), ch = getchar();
  return f ? -x : x;
}
const int N = 4e5 + 10;
int n, m, incf[N], pref[N], p[N], s[N], b[N];
int a[N], tot, cnt, ans[N];
int pos(int x) { return lower_bound(a + 1, a + tot + 1, x) - a; }
struct line {
  int ti, x, y, op, id;
} l[N];
struct BIT {
  int tr[N];
  void add(int x, int y) {
    for (; x <= tot; x += x & -x) tr[x] += y;
  }
  int query(int x, int y = 0) {
    for (; x; x -= x & -x) y += tr[x];
    return y;
  }
} A, B;
int main() {
  n = read(), m = read();
  for (int i = 1; i <= n; i++) p[i] = read();
  for (int i = 1; i <= n; i++) s[i] = read();
  for (int i = 1; i <= n; i++) b[i] = read();
  for (int i = 1; i <= m; i++) incf[i] = read();
  for (int i = 1; i <= m; i++) pref[i] = read();
  for (int i = 1; i <= n; i++)
    a[++tot] = b[i] + p[i], a[++tot] = b[i] - p[i] + 1;
  for (int i = 1; i <= m; i++)
    a[++tot] = incf[i] + pref[i], a[++tot] = pref[i] - incf[i];
  sort(a + 1, a + tot + 1);
  tot = unique(a + 1, a + tot + 1) - a - 1;
  for (int i = 1; i <= n; i++)
    l[++cnt] = (line){p[i], p[i], b[i], 1, 0},
    l[++cnt] = (line){s[i] + 1, p[i], b[i], -1, 0};
  for (int i = 1; i <= m; i++)
    l[++cnt] = (line){incf[i], incf[i], pref[i], 2, i};
  sort(l + 1, l + cnt + 1, [](line a, line b) {
    return a.ti == b.ti ? a.id == b.id ? a.op > b.op : a.op < b.op
                        : a.ti < b.ti;
  });
  for (int i = 1; i <= cnt; i++) {
    if (l[i].id)
      ans[l[i].id] =
          A.query(pos(l[i].x + l[i].y)) - B.query(pos(l[i].y - l[i].x));
    else
      A.add(pos(l[i].x + l[i].y), l[i].op),
          B.add(pos(l[i].y - l[i].x + 1), l[i].op);
  }
  for (int i = 1; i <= m; i++) printf("%d%c", ans[i], " \n"[i == m]);
  return 0;
}