Polycarp has n coins, the value of the i-th coin is a_i. It is guaranteed that all the values are integer powers of 2 (i.e. a_i = 2^d for some non-negative integer number d).

Polycarp wants to know answers on q queries. The j-th query is described as integer number b_j. The answer to the query is the minimum number of coins that is necessary to obtain the value b_j using some subset of coins (Polycarp can use only coins he has). If Polycarp can't obtain the value b_j, the answer to the j-th query is -1.

The queries are independent (the answer on the query doesn't affect Polycarp's coins).

Input

The first line of the input contains two integers n and q (1 ≤ n, q ≤ 2 ⋅ 10^5) — the number of coins and the number of queries.

The second line of the input contains n integers a_1, a_2, ..., a_n — values of coins (1 ≤ a_i ≤ 2 ⋅ 10^9). It is guaranteed that all a_i are integer powers of 2 (i.e. a_i = 2^d for some non-negative integer number d).

The next q lines contain one integer each. The j-th line contains one integer b_j — the value of the j-th query (1 ≤ b_j ≤ 10^9).

Output

Print q integers ans_j. The j-th integer must be equal to the answer on the j-th query. If Polycarp can't obtain the value b_j the answer to the j-th query is -1.

Example

Input

5 4
2 4 8 2 4
8
5
14
10


Output

1
-1
3
2
#include <bits/stdc++.h>
using namespace std;
long long solve(long long n) {
  long long cnt = 0;
  while (n % 2 == 0) {
    cnt++;
    n /= 2;
  }
  return cnt;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  vector<long long> arr;
  vector<long long> freq(40, 0);
  long long p = 1;
  for (int i = 0; i < 40; i++) {
    arr.push_back(p);
    p *= 2;
  }
  long long n, q, x, num;
  long long sum = 0;
  cin >> n >> q;
  for (int i = 0; i < n; i++) {
    cin >> x;
    sum += x;
    freq[solve(x)]++;
  }
  while (q--) {
    cin >> num;
    long long hi = 39;
    long long cnt = 0;
    while (hi >= 0) {
      if (arr[hi] > num) {
        hi--;
        continue;
      } else {
        long long p = num / arr[hi];
        if (p >= freq[hi]) {
          num -= (freq[hi] * arr[hi]);
          cnt += freq[hi];
        } else {
          num -= (p * arr[hi]);
          cnt += p;
        }
        hi--;
      }
    }
    if (num > 0)
      cout << -1;
    else
      cout << cnt;
    cout << "\n";
  }
}
