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;
template <typename T, typename U>
std::pair<T, U> operator+(const std::pair<T, U>& l, const std::pair<T, U>& r) {
  return {l.first + r.first, l.second + r.second};
}
typedef void (*callback_function)(void);
const long long ZERO = 0LL;
const long long INF64 = 1e18;
const long long INF32 = 1e9;
const long long MOD = 1e9 + 7;
const long double PI = acos(-1.0L);
const long double EPS = static_cast<long double>(1e-9);
const long long N = 1e6 + 7;
long long n, q;
long long a[N], b[N], res[N];
map<long long, long long> mp;
void input() {
  cin >> n >> q;
  for (long long i = (0); i < (n); i++) {
    cin >> a[i];
  }
  for (long long i = (0); i < (q); i++) {
    cin >> b[i];
  }
}
void debug() {
  if (true) {
  }
}
void solve() {
  for (long long i = (0); i < (n); i++) {
    if (!mp.count(a[i])) {
      mp[a[i]] = 0;
    }
    mp[a[i]]++;
  }
  for (long long i = (0); i < (q); i++) {
    map<long long, long long> mp2 = mp;
    for (long long x = 1LL << 32; x > 0; x /= 2) {
      while (x & b[i]) {
        long long y = x;
        while (y > 0 and !mp2.count(y)) {
          y /= 2;
        }
        if (y == 0) {
          res[i] = -1;
          break;
        }
        long long cnt = min(x / y, mp2[y]);
        res[i] += cnt;
        mp2[y] -= cnt;
        if (mp2[y] == 0) {
          mp2.erase(y);
        }
        b[i] -= y * cnt;
      }
    }
  }
}
void output() {
  for (long long i = (0); i < (q); i++) {
    cout << res[i] << endl;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie();
  {
    input();
    solve();
    output();
  }
  return 0;
}
