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 long long INF = 1e9;
struct segTree {
  int n;
  vector<long long> tree, lazy;
  segTree(int n) : tree(4 * n, 0), lazy(4 * n, 0), n(n) {}
  void propagate(int node, int l, int r) {
    tree[node] += lazy[node] * (r - l + 1);
    if (l != r) lazy[2 * node] += lazy[node], lazy[2 * node + 1] += lazy[node];
    lazy[node] = 0;
  }
  void treeUpdate(int node, int l, int r, int u, int v, long long val) {
    int mid = (l + r) >> 1;
    propagate(node, l, r);
    if (l > v || r < u) return;
    if (l >= u && r <= v) {
      lazy[node] += val;
      propagate(node, l, r);
      return;
    }
    treeUpdate(2 * node, l, mid, u, v, val);
    treeUpdate(2 * node + 1, mid + 1, r, u, v, val);
    tree[node] = tree[2 * node] + tree[2 * node + 1];
  }
  void update(int l, int r, long long val) { treeUpdate(1, 1, n, l, r, val); }
  long long treeQuery(int node, int l, int r, int u, int v) {
    int mid = (l + r) >> 1;
    propagate(node, l, r);
    if (l > v || r < u) return 0;
    if (l >= u && r <= v) return tree[node];
    return treeQuery(2 * node, l, mid, u, v) +
           treeQuery(2 * node + 1, mid + 1, r, u, v);
  }
  long long query(int l, int r) { return treeQuery(1, 1, n, l, r); }
};
struct person {
  int inc, pref, idx;
  person() {}
  person(int inc, int pref, int idx) : inc(inc), pref(pref), idx(idx) {}
  bool operator<(const person& rhs) const { return inc < rhs.inc; }
} ara[100005];
struct item {
  int p, s, b;
} items[100005];
bool cmp1(item a, item b) { return a.p + a.b < b.p + b.b; }
bool cmp2(item a, item b) { return a.p - a.b < b.p - b.b; }
int ans[100005];
int main() {
  int n, m;
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", &items[i].p);
  for (int i = 1; i <= n; i++) scanf("%d", &items[i].s);
  for (int i = 1; i <= n; i++) scanf("%d", &items[i].b);
  for (int i = 1; i <= m; i++) scanf("%d", &ara[i].inc);
  for (int i = 1; i <= m; i++) scanf("%d", &ara[i].pref);
  for (int i = 1; i <= m; i++) ara[i].idx = i;
  sort(ara + 1, ara + m + 1);
  segTree tree(m);
  tree.update(1, m, -INF);
  sort(items + 1, items + n + 1, cmp1);
  priority_queue<pair<int, int>, vector<pair<int, int> >,
                 greater<pair<int, int> > >
      pq;
  for (int i = 1; i <= m; i++) pq.push({ara[i].inc + ara[i].pref, i});
  for (int i = 1; i <= n; i++) {
    while (!pq.empty()) {
      pair<int, int> cur = pq.top();
      if (cur.first >= items[i].b + items[i].p) break;
      pq.pop();
      int j = cur.second;
      int lagbe = tree.query(j, j);
      tree.update(j, j, -lagbe);
    }
    int p = items[i].p, b = items[i].b, s = items[i].s;
    int x = lower_bound(ara + 1, ara + m + 1, person(p, 0, 0)) - ara;
    int y = upper_bound(ara + 1, ara + m + 1, person(s, 0, 0)) - ara - 1;
    if (x > y) continue;
    tree.update(x, y, 1);
  }
  for (int i = 1; i <= m; i++) ans[ara[i].idx] += max(0LL, tree.query(i, i));
  tree = segTree(m);
  tree.update(1, m, -INF);
  sort(items + 1, items + n + 1, cmp2);
  while (!pq.empty()) pq.pop();
  for (int i = 1; i <= m; i++) pq.push({ara[i].inc - ara[i].pref, i});
  for (int i = 1; i <= n; i++) {
    while (!pq.empty()) {
      pair<int, int> cur = pq.top();
      if (cur.first >= items[i].p - items[i].b) break;
      pq.pop();
      int j = cur.second;
      int lagbe = tree.query(j, j);
      tree.update(j, j, -lagbe);
    }
    int p = items[i].p, b = items[i].b, s = items[i].s;
    int x = lower_bound(ara + 1, ara + m + 1, person(p, 0, 0)) - ara;
    int y = upper_bound(ara + 1, ara + m + 1, person(s, 0, 0)) - ara - 1;
    if (x > y) continue;
    tree.update(x, y, 1);
  }
  for (int i = 1; i <= m; i++) ans[ara[i].idx] += max(0LL, tree.query(i, i));
  tree = segTree(m);
  for (int i = 1; i <= n; i++) {
    int p = items[i].p, b = items[i].b, s = items[i].s;
    int x = lower_bound(ara + 1, ara + m + 1, person(p, 0, 0)) - ara;
    int y = upper_bound(ara + 1, ara + m + 1, person(s, 0, 0)) - ara - 1;
    if (x > y)
      tree.update(1, m, 1);
    else {
      if (x > 1) tree.update(1, x - 1, 1);
      if (y < m) tree.update(y + 1, m, 1);
    }
  }
  for (int i = 1; i <= m; i++) ans[ara[i].idx] += max(0LL, tree.query(i, i));
  for (int i = 1; i <= m; i++) ans[i] = n - ans[i];
  for (int i = 1; i <= m; i++) printf("%d ", ans[i]);
  return 0;
}