codeforces_632E.Thief in a Shop（dp）

E. Thief in a Shop

time limit per test

5 seconds

memory limit per test

512 megabytes

input

standard input

output

standard output

A thief made his way to a shop.

As usual he has his lucky knapsack with him. The knapsack can contain k objects. There are n kinds of products in the shop and an infinite number of products of each kind. The cost of one product of kind i is ai.

The thief is greedy, so he will take exactly k products (it's possible for some kinds to take several products of that kind).

Find all the possible total costs of products the thief can nick into his knapsack.

Input

The first line contains two integers n and k (1 ≤ n, k ≤ 1000) — the number of kinds of products and the number of products the thief will take.

The second line contains n integers ai (1 ≤ ai ≤ 1000) — the costs of products for kinds from 1 to n.

Output

Print the only line with all the possible total costs of stolen products, separated by a space. The numbers should be printed in the ascending order.

Examples

input

3 2
1 2 3

output

2 3 4 5 6

input

5 5
1 1 1 1 1

output

5

input

3 3
3 5 11

output

9 11 13 15 17 19 21 25 27 33

好久每做dp了，都没思路了，正好选k个想不到怎么dp，于是让n个ai都减去a1，使得第一个数是0，后面不足k个再加上k*a1。

#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;

int dp[1000010];

int main()
{
    int i,j,n,k,a[1010];
    while(scanf("%d%d",&n,&k)!=EOF)
    {
        for(i=1; i<=n; i++)scanf("%d",&a[i]);
        sort(a+1,a+n+1);
        int a1=a[1];
        for(i=1; i<=n; i++)a[i]-=a1;
        for(i=1; i<=1000000; i++)dp[i]=99999999;
        dp[0]=0;
        for(i=0; i<=1000000; i++)
        {
            for(j=1; j<=n; j++)
            {
                if(i>=a[j])
                {
                    dp[i]=min(dp[i],dp[i-a[j]]+1);
                }
            }
        }
        for(i=0; i<=1000000; i++)
        {
            if(dp[i]<=k)
            {
                printf("%d ",i+k*a1);
            }
        }
        printf("\n");
    }


    return 0;
}