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 mod = 1000000007;
long long gcd(long long a, long long b) { return (b == 0) ? a : gcd(b, a % b); }
int main() {
  ios_base::sync_with_stdio(false);
  long long n;
  cin >> n;
  long long q;
  cin >> q;
  long long sum = 0;
  vector<long long> v;
  vector<long long> count(40, 0);
  for (int i = 0; i < n; i++) {
    long long a;
    cin >> a;
    v.push_back(a);
    long long x = log2(a);
    count[x]++;
    sum += a;
  }
  while (q--) {
    long long a;
    cin >> a;
    if (a > sum) {
      cout << -1 << endl;
      continue;
    }
    long long x = a;
    int c = 0;
    int flag = 0;
    vector<long long> op = count;
    int found = 0;
    int hk = 0;
    while (x > 0) {
      if (x & 1) {
        found = 0;
        if (op[c] >= 1) {
          hk++;
          found = 1;
          op[c]--;
        } else {
          int j = 1;
          for (int i = c; i >= 0; i--) {
            if (op[i] >= j) {
              found = 1;
              op[i] -= j;
              hk += j;
              break;
            } else {
              hk += op[i];
              j -= op[i];
            }
            j = j * 2;
          }
          if (j == 0) {
            found = 1;
          }
          if (found == 0) {
            flag = 1;
            break;
          }
        }
        if (found == 0) {
          flag = 1;
          break;
        }
      }
      c++;
      x = x >> 1;
    }
    if (flag) {
      cout << -1 << endl;
    } else {
      cout << hk << endl;
    }
  }
  return 0;
}
