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 = 1e5 + 10;
int P[maxn], S[maxn], B[maxn], INC[maxn], PREF[maxn], MI[maxn];
int stn[maxn], stm[maxn], ans[maxn], res[maxn * 2], tmp[maxn * 2];
int n, m;
bool cmp_S(int a, int b) { return S[a] > S[b]; }
bool cmp_INC(int a, int b) { return INC[a] > INC[b]; }
bool cmp_res(int a, int b) {
  int vala = a <= n ? P[a] : PREF[a - n];
  int valb = b <= n ? P[b] : PREF[b - n];
  if (vala != valb) return vala < valb;
  return a <= n && b > n;
}
struct BIT {
  int bit[maxn * 2];
  void add(int x) {
    while (x < maxn * 2) {
      bit[x]++;
      x += x & -x;
    }
  }
  int sum(int x) {
    int ret = 0;
    while (x > 0) {
      ret += bit[x];
      x -= x & -x;
    }
    return ret;
  }
  void init() { memset(bit, 0, sizeof(bit)); }
} tree;
void binary(int nl, int nr, int ml, int mr, int tl, int tr) {
  if (nl > nr) {
    for (int i = tl; i <= tr; i++) res[i] = n + stm[ml + i - tl];
    sort(res + tl, res + tr + 1, cmp_res);
    return;
  }
  if (ml > mr) {
    for (int i = tl; i <= tr; i++) res[i] = stn[nl + i - tl];
    sort(res + tl, res + tr + 1, cmp_res);
    return;
  }
  if (nl == nr) {
    for (int i = ml; i <= mr; i++) tmp[i] = stm[i] + n;
    sort(tmp + ml, tmp + mr + 1, cmp_res);
    int mpos = ml, pos = tl;
    while (mpos <= mr && PREF[tmp[mpos] - n] < P[stn[nl]]) {
      res[pos++] = tmp[mpos++];
    }
    res[pos++] = stn[nl];
    while (mpos <= mr) {
      if (B[stn[nl]] <= MI[tmp[mpos] - n]) ans[tmp[mpos] - n]++;
      res[pos++] = tmp[mpos++];
    }
    return;
  }
  int nmid = (nl + nr) / 2;
  int mmid = ml - 1;
  while (mmid + 1 <= mr && INC[stm[mmid + 1]] > S[stn[nmid + 1]])
    mmid = mmid + 1;
  binary(nl, nmid, ml, mmid, tl, tl + nmid - nl + 1 + mmid - ml + 1 - 1);
  binary(nmid + 1, nr, mmid + 1, mr, tl + nmid - nl + 1 + mmid - ml + 1, tr);
  int lpos = tl, rpos = tl + nmid - nl + 1 + mmid - ml + 1;
  tree.init();
  for (int i = tl; i <= tr; i++) {
    if (rpos > tr || lpos <= tl + nmid - nl + 1 + mmid - ml + 1 - 1 &&
                         cmp_res(res[lpos], res[rpos])) {
      if (res[lpos] <= n) {
        tree.add(B[res[lpos]]);
      }
      tmp[i] = res[lpos++];
    } else {
      if (res[rpos] > n) {
        ans[res[rpos] - n] += tree.sum(MI[res[rpos] - n]);
      }
      tmp[i] = res[rpos++];
    }
  }
  for (int i = tl; i <= tr; i++) res[i] = tmp[i];
}
vector<int> lsh;
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 <= n; i++) {
    int tmp = P[i];
    P[i] = tmp + B[i];
    B[i] = tmp - B[i];
  }
  for (int i = 1; i <= m; i++) {
    MI[i] = INC[i] - PREF[i];
    PREF[i] += INC[i];
  }
  for (int i = 1; i <= n; i++) lsh.push_back(B[i]);
  for (int i = 1; i <= m; i++) lsh.push_back(MI[i]);
  sort(lsh.begin(), lsh.end());
  lsh.erase(unique(lsh.begin(), lsh.end()), lsh.end());
  for (int i = 1; i <= n; i++)
    B[i] = lower_bound(lsh.begin(), lsh.end(), B[i]) - lsh.begin() + 1;
  for (int i = 1; i <= m; i++)
    MI[i] = lower_bound(lsh.begin(), lsh.end(), MI[i]) - lsh.begin() + 1;
  for (int i = 1; i <= n; i++) stn[i] = i;
  sort(stn + 1, stn + 1 + n, cmp_S);
  for (int i = 1; i <= m; i++) stm[i] = i;
  sort(stm + 1, stm + 1 + m, cmp_INC);
  int stpos = 1;
  while (stpos <= m && INC[stm[stpos]] > S[stn[1]]) stpos++;
  binary(1, n, stpos, m, 1, n + m - stpos + 1);
  for (int i = 1; i <= m; i++) printf("%d%c", ans[i], i == m ? '\n' : ' ');
  return 0;
}