E. Thief in a Shop

 

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

 

 

题意：

　　给你n个数，你可以选数选k次，每次可以选相同的数，问你有多少种和

题解：

　　将k分解

　　进行log(k)次FFT

　　FFT精度可能不够，bool好像能过

　　NTT，换一个合适 的 费马素数和原格就能过

 　　复杂度：O(n*log(n)*log(k))

#include
             
              using namespace std;#pragma comment(linker, "/STACK:102400000,102400000")#define ls i<<1#define rs ls | 1#define mid ((ll+rr)>>1)#define pii pair
              
               #define MP make_pairtypedef long long LL;const long long INF = 1e18+1LL;const double pi = acos(-1.0);const int N = 5e5+10, M = 1e3+20,inf = 2e9,mod = 1e9+7;const long long P=23068673LL;const int G=3;LL mul(LL x,LL y){    return (x*y-(LL)(x/(long double)P*y+1e-3)*P+P)%P;}LL qpow(LL x,LL k,LL p){    LL ret=1;    while(k){        if(k&1) ret=mul(ret,x);        k>>=1;        x=mul(x,x);    }    return ret;}LL wn[50];void getwn(){    for(int i=1; i<=22; ++i){        int t=1<
               
                >1,k; i
                
                 >1;        while(j>=k){            j-=k;            k>>=1;        }        if(j
                 
                  >1); ++j){                LL u=y[j],t=mul(y[j+h/2],w);                y[j]=u+t;                if(y[j]>=P) y[j]-=P;                y[j+h/2]=u-t+P;                if(y[j+h/2]>=P) y[j+h/2]-=P;                w=mul(w,wn[id]);            }        }    }    if(op==-1){        for(int i=1; i
                  
                    0) { if(m%2) { Convolution(now,num); } Convolution(num,num); m/=2; } NTT(now,-1); for(int i = 1; i <= mx; ++i) { if(now[i]) printf("%d ",i); } return 0;}/*3 33 5 11*/