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 power(long long x, long long y, long long p) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
int main() {
  long long n, q;
  cin >> n >> q;
  long long a[n];
  map<long long, long long> fre;
  map<long long, long long, greater<long long>> pwr;
  for (long long i = 0; i < n; i++) cin >> a[i], fre[a[i]]++, pwr[log2(a[i])]++;
  sort(a, a + n);
  long long b = 0;
  while (q--) {
    cin >> b;
    if (b % 2 == 1 && fre[1] == 0)
      cout << -1 << endl;
    else {
      long long low = 0, high = n - 1;
      long long flag = 0, lb = 0;
      while (low <= high) {
        long long mid = low + (high - low) / 2;
        if (a[mid] < b) {
          lb = max(lb, a[mid]);
          low = mid + 1;
        } else if (a[mid] > b) {
          high = mid - 1;
        } else {
          flag = 1;
          break;
        }
      }
      if (flag)
        cout << 1 << endl;
      else {
        long long cnt = 0, mx = log2(lb);
        for (auto itr = pwr.find(mx); itr != pwr.end(); itr++) {
          long long ex = itr->first;
          long long val = pow(2, ex);
          long long ini = b;
          b -= min((b / val) * val, pwr[ex] * val);
          cnt += (ini - b) / val;
        }
        if (b == 0)
          cout << cnt << endl;
        else
          cout << -1 << endl;
      }
    }
  }
  return 0;
}
