map首例

Polycarp has nn coins, the value of the ii-th coin is aiai. It is guaranteed that all the values are integer powers of 22 (i.e. ai=2dai=2d for some non-negative integer number dd).

Polycarp wants to know answers on qq queries. The jj-th query is described as integer number bjbj. The answer to the query is the minimum number of coins that is necessary to obtain the value bjbj using some subset of coins (Polycarp can use only coins he has). If Polycarp can't obtain the value bjbj, the answer to the jj-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 nn and qq (1≤n,q≤2⋅1051≤n,q≤2⋅105) — the number of coins and the number of queries.

The second line of the input contains nn integers a1,a2,…,ana1,a2,…,an — values of coins (1≤ai≤2⋅1091≤ai≤2⋅109). It is guaranteed that all aiai are integer powers of 22 (i.e. ai=2dai=2dfor some non-negative integer number dd).

The next qq lines contain one integer each. The jj-th line contains one integer bjbj — the value of the jj-th query (1≤bj≤1091≤bj≤109).

Output

Print qq integers ansjansj. The jj-th integer must be equal to the answer on the jj-th query. If Polycarp can't obtain the value bjbj the answer to the jj-th query is -1.

Example

Input

5 4
2 4 8 2 4
8
5
14
10

Output

1
-1
3
2

该题由于数据量过大，用map记录每个2的n次方出现的个数

 

 

#include <bits/stdc++.h>
using namespace std;
const int MAXN = 2e5+5;
map<int, int> mp;
int n, q;
int main()
{
    scanf("%d%d", &n, &q);
    for (int i = 0; i < n; i++)
    {
        int t;
        scanf("%d", &t);
        mp[t]++;
    }
    while (q--)
    {
        int b;
        scanf("%d", &b);
        int ans = 0;
        for (int i = 1<<30; i >= 1; i /= 2)
        {
            int x = min(mp[i], b/i);
            ans += x;
            b -= x*i;
        }
        if (b)
            ans = -1;
        printf("%d\n", ans);
    }
    int p=1<<30;
    printf("%d\n",p);
    return 0;
}