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 maxn = 400010;
int n, m, p[maxn], s[maxn], b[maxn], inc[maxn], pre[maxn], ans[maxn], top;
int A[maxn], B[maxn], cnt1, cnt2, c1[maxn], c2[maxn];
struct node {
  int id, x, y;
} e[maxn];
bool cmp(node a, node b) {
  if (a.x == b.x) return a.id < b.id;
  return a.x < b.x;
}
int query1(int x) {
  int res = 0;
  while (x) {
    res += c1[x];
    x -= x & (-x);
  }
  return res;
}
void add1(int x, int v) {
  while (x <= cnt1) {
    c1[x] += v;
    x += x & (-x);
  }
}
int query2(int x) {
  int res = 0;
  while (x) {
    res += c2[x];
    x -= x & (-x);
  }
  return res;
}
void add2(int x, int v) {
  while (x <= cnt2) {
    c2[x] += v;
    x += x & (-x);
  }
}
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", &pre[i]);
  for (int i = 1; i <= n; i++) {
    A[++cnt1] = b[i] + p[i];
    B[++cnt2] = b[i] - p[i] + 1;
    ++top;
    e[top].x = p[i];
    e[top].y = i;
    e[top].id = 0;
    ++top;
    e[top].x = s[i] + 1;
    e[top].y = i;
    e[top].id = -1;
  }
  for (int i = 1; i <= m; i++) {
    A[++cnt1] = inc[i] + pre[i];
    B[++cnt2] = pre[i] - inc[i];
    ++top;
    e[top].x = inc[i];
    e[top].y = i;
    e[top].id = i;
  }
  sort(A + 1, A + 1 + cnt1);
  cnt1 = unique(A + 1, A + 1 + cnt1) - A - 1;
  sort(B + 1, B + 1 + cnt2);
  cnt2 = unique(B + 1, B + 1 + cnt2) - B - 1;
  sort(e + 1, e + 1 + top, cmp);
  for (int i = 1; i <= top; i++) {
    if (e[i].id == 0) {
      int temp = e[i].y;
      add1(lower_bound(A + 1, A + 1 + cnt1, b[temp] + p[temp]) - A, 1);
      add2(lower_bound(B + 1, B + 1 + cnt2, b[temp] - p[temp] + 1) - B, 1);
    } else if (e[i].id == -1) {
      int temp = e[i].y;
      add1(lower_bound(A + 1, A + 1 + cnt1, b[temp] + p[temp]) - A, -1);
      add2(lower_bound(B + 1, B + 1 + cnt2, b[temp] - p[temp] + 1) - B, -1);
    } else {
      int temp = e[i].y;
      ans[e[i].id] =
          query1(lower_bound(A + 1, A + 1 + cnt1, inc[temp] + pre[temp]) - A);
      ans[e[i].id] -=
          query2(lower_bound(B + 1, B + 1 + cnt2, pre[temp] - inc[temp]) - B);
    }
  }
  for (int i = 1; i <= m; i++) printf("%d ", ans[i]);
  return 0;
}