## Algorithm

D. Coins and Queries
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

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

Polycarp wants to know answers on q queries. The j-th query is described as integer number bj��. The answer to the query is the minimum number of coins that is necessary to obtain the value bj�� using some subset of coins (Polycarp can use only coins he has). If Polycarp can't obtain the value bj��, 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 (1n,q21051≤�,�≤2⋅105) — the number of coins and the number of queries.

The second line of the input contains n integers a1,a2,,an�1,�2,…,�� — values of coins (1ai21091≤��≤2⋅109). It is guaranteed that all ai�� are integer powers of 22 (i.e. ai=2d��=2� for some non-negative integer number d).

The next q lines contain one integer each. The j-th line contains one integer bj�� — the value of the j-th query (1bj1091≤��≤109).

Output

Print q integers ansj����. The j-th integer must be equal to the answer on the j-th query. If Polycarp can't obtain the value bj�� the answer to the j-th query is -1.

Example
input
Copy
5 42 4 8 2 4851410
output
Copy
1-132

## Code Examples

### #1 Code Example with C++ Programming

Code - C++ Programming

#include <bits/stdc++.h>

using namespace std;

int const N = 2e5 + 1;
int n, q, a[N], fr[35], fr1[35];

bool check() {
for(int i = 0; i < 35; ++i)
if(fr1[i] != 0)
return false;
return true;
}

int main() {
scanf("%d %d", &n, &q);
for(int i = 0; i < n; ++i) {
scanf("%d", a + i);
bitset<32> bs(a[i]);
for(int i = 0; i < 32; ++i)
if(bs[i] == 1)
++fr[i];
}

while(q-- != 0) {
int tmp;
scanf("%d", &tmp>;
bitset<32> bs(tmp);

memset(fr1, 0, sizeof fr1);
for(int i = 0; i < 32; ++i)
fr1[i] = bs[i];

int cnt = 0;
for(int i = 31; i >= 0; --i) {
if(i > 0 && fr1[i] != 0 && fr[i] == 0) {
fr1[i - 1] += fr1[i] * 2;
fr1[i] = 0;
} else if(i > 0 && fr1[i] != 0 && fr[i] < fr1[i]) {
fr1[i - 1] += (fr1[i] - fr[i]) * 2;
cnt += fr[i];
fr1[i] = 0;
} else if(fr1[i] != 0 && fr[i] >= fr1[i]) {
cnt += fr1[i];
fr1[i] = 0;
}
}

if(check())
printf("%d\n", cnt);
else
puts("-1");
}

return 0;
}
Copy The Code &

Input

cmd
5 4
2 4 8 2 4
8
5
14
10

Output

cmd
1
-1
3
2