CodeForces - 1003D

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=2d for 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

 1 #include<cstdio>
 2 #include<cstdlib>
 3 #include<cstring>
 4 #include<cmath>
 5 #include<algorithm>
 6 #include<queue>
 7 #include<stack>
 8 #include<deque>
 9 #include<map>
10 #include<iostream>
11 using namespace std;
12 typedef long long  LL;
13 const double pi=acos(-1.0);
14 const double e=exp(1);
15 map<long long,long long> mp;
16 int main()
17 {
18     LL i,p,j,n,q;
19     LL ans;
20     scanf("%lld%lld",&n,&q);
21     for(i=1;i<=n;i++)
22     {
23         scanf("%lld",&p);
24         mp[p]++;
25     }
26     while(q--)
27     {
28         LL x,mid;
29         ans=0;
30         scanf("%lld",&x);
31         for(i=1<<30;i>=1;i/=2)
32         {
33             mid=min(mp[i],x/i);
34             ans+=mid;
35             x-=mid*i;
36         }
37         if(x!=0)
38             printf("-1\n");
39         else
40             printf("%lld\n",ans);
41     }
42     return 0;
43 }
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posted @ 2018-07-29 16:40  Daybreaking  阅读(205)  评论(0)  编辑  收藏  举报