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
//Thief in a Shop
import java.io.*;
import java.util.*;

public class E632{
   
   public static void main(String[] args)throws IOException{
      BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
      PrintWriter out = new PrintWriter(System.out);
      
      StringTokenizer st = new StringTokenizer(f.readLine());
      
      int n = Integer.parseInt(st.nextToken());
      int m = Integer.parseInt(st.nextToken());
      
      long[] array = new long[1001];
      st = new StringTokenizer(f.readLine());
      for(int k = 0; k < n; k++){
         int i = Integer.parseInt(st.nextToken());
         array[i] = 1L;
      }
      
      long[] answer = polyexp(array,m);
      
      for(int k = 1; k < answer.length; k++){
         if(answer[k] > 0L) out.print(k + " ");
      }
      
      
      
      
      
      
      out.close();
   }
   
   public static long[] polyexp(long[] array, int power){
      if(power == 0) 
         return new long[]{1L};
      if(power == 1) 
         return array;
      
      long[] ans = new long[]{1L};
      int p = power;
      while(p > 0){
         if(p%2 == 1) ans = mul(ans,array);
         p >>= 1;
         array = mul(array,array);
         //for(int k = 0; k < ans.length; k++) ans[k] = Math.min(ans[k],1);                 //MAKES EVERYTHING 0 OR 1
      }
      
      
      return ans;
   }
   
   //FFT TEMPLATE
   //from https://judge.yosupo.jp/submission/8499
   public static final long p=998_244_353;                       //mod
   public static final long g=3;                                 //primitive root
	
   public static long ADD(long a,long b) {
      return a+b>=p?a+b-p:a+b;
   }
	
   public static long SUB(long a,long b) {
      return ADD(a,p-b);
   }
	
   public static long pow(long a,long n) {
      long ret=1;
      for (;n>0;n/=2,a=a*a%p) 
         if(n%2==1) ret=ret*a%p;
      return ret;
   }
	
   public static long inv(long a) {
      return pow(a,p-2);
   }
	
   public static void fft(long[] a,long root) {
      int n=a.length;
      {
         int cur=0;
         for (int i=0;i<n;++i) {
            if (i<cur) {
               a[i]^=a[cur];a[cur]^=a[i];a[i]^=a[cur];
            }
            for (int k=n>>1;k>(cur^=k);k>>=1) ;
         }
      }
   	
      for (int s=1;s<=n/2;s*=2) {
         long mul=pow(root,n/(2*s));
         for (int i=0;i<n;i+=2*s) {
            long x=1;
            for (int j=0;j<s;++j) {
               long A=a[i+j];
               long B=a[i+j+s]*x%p;
               x=x*mul%p;
               a[i+j]=ADD(A,B);
               a[i+j+s]=SUB(A,B);
            }
         }
      }
   }
	
   public static long[] mul(long[] a,long[] b) {
      int n=1;
      while (n<a.length+b.length-1) n*=2;
      a=Arrays.copyOf(a, n);
      b=Arrays.copyOf(b, n);
      long root=pow(g,(p-1)/n);
      fft(a,root);
      fft(b,root);
      long in=inv(n);
      for (int i=0;i<n;++i) a[i]=a[i]*b[i]%p*in%p;
      fft(a,inv(root));
      
      for(int k = 0; k < a.length; k++) a[k] = Math.min(a[k],1);
      return a;
   }
      
}