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;
const double EPS = 1e-9;
const double pi = acos(-1);
int dx[] = {0, 0, 1, -1, 1, 1, -1, -1};
int dy[] = {1, -1, 0, 0, -1, 1, 1, -1};
const int N = 55;
vector<long long> adj[55];
int main() {
  int n, q;
  cin >> n >> q;
  int a[n];
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    int indx = (int)log2(a[i]);
    adj[indx].push_back(a[i]);
  }
  for (int i = 0; i < N; ++i)
    if (!adj[i].empty())
      partial_sum((adj[i]).begin(), (adj[i]).end(), adj[i].begin());
  int num;
  while (q--) {
    cin >> num;
    int cnt = 0;
    for (int i = N - 1; ~i; --i) {
      if (!adj[i].empty()) {
        auto it = upper_bound((adj[i]).begin(), (adj[i]).end(), num);
        int indx = it - adj[i].begin();
        cnt += indx;
        if (indx) num -= *(--it);
      }
    }
    if (num)
      puts("-1");
    else
      cout << cnt << endl;
  }
}
