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 N = 1e5 + 5;
int p[N], s[N], b[N], inc[N], pref[N];
int n, m;
struct Node {
  int x, y, id;
  Node(int x = 0, int y = 0, int id = 0) : x(x), y(y), id(id) {}
  bool operator<(const Node b) const { return x < b.x; }
} nd[N], Inc[N], Dec[N];
int L, R;
const long long P = 2e9;
struct SegT {
  int root;
  int t[N * 32], cnt, lch[N * 32], rch[N * 32];
  void add(int &x, long long l, long long r, long long pos, int sum) {
    if (!x) x = ++cnt;
    t[x] += sum;
    long long mid = (l + r) >> 1;
    if (l == r) return;
    if (pos <= mid)
      add(lch[x], l, mid, pos, sum);
    else
      add(rch[x], mid + 1, r, pos, sum);
  }
  int qry(int x, long long l, long long r, long long ql, long long qr) {
    if (!x) return 0;
    long long mid = (l + r) >> 1;
    int ans = 0;
    if (ql <= l && qr >= r) return t[x];
    if (ql <= mid) ans += qry(lch[x], l, mid, ql, qr);
    if (qr > mid) ans += qry(rch[x], mid + 1, r, ql, qr);
    return ans;
  }
} T1, T2;
int ret[N];
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", &p[i]);
  for (int i = 1; i <= n; i++) scanf("%d", &s[i]);
  for (int i = 1; i <= n; i++) scanf("%d", &b[i]);
  for (int i = 1; i <= m; i++) scanf("%d", &inc[i]);
  for (int i = 1; i <= m; i++) scanf("%d", &pref[i]);
  for (int i = 1; i <= m; i++) nd[i] = Node(inc[i], pref[i], i);
  for (int i = 1; i <= n; i++) {
    Inc[i] = Node(p[i], b[i], i);
    Dec[i] = Node(s[i], b[i], i);
  }
  sort(nd + 1, nd + m + 1);
  sort(Inc + 1, Inc + n + 1);
  sort(Dec + 1, Dec + n + 1);
  int ans = 0;
  for (int i = 1; i <= m; i++) {
    int X = nd[i].x, Y = nd[i].y;
    while (L + 1 <= n && Inc[L + 1].x <= X) {
      ++L, ++ans;
      long long xx = p[Inc[L].id], yy = b[Inc[L].id];
      T1.add(T1.root, -P, P, xx + yy, 1);
      T2.add(T2.root, -P, P, yy - xx, 1);
    }
    while (R + 1 <= n && Dec[R + 1].x < X) {
      ++R, --ans;
      long long xx = p[Dec[R].id], yy = b[Dec[R].id];
      T1.add(T1.root, -P, P, xx + yy, -1);
      T2.add(T2.root, -P, P, yy - xx, -1);
    }
    int D = 0;
    D += T1.qry(T1.root, -P, P, X + Y + 1, P);
    D += T2.qry(T2.root, -P, P, -P, Y - X - 1);
    ret[nd[i].id] = ans - D;
  }
  for (int i = 1; i <= m; i++) printf("%d ", ret[i]);
}