The Values You Can Make

对于一个集合它的子集的个数为2^n（对于一个有n个元素的集合，一个元素存在和不存在子集中就有2种不同的子集，所以n个元素就是2^n个不同子集）

dp[i][j] i表示现在的金额，j表示构成现在金额的钱的一个子集

dp[i][j+a]表示a这个元素在这个金额i的子集中

反之，dp[i][j]表示a这个元素不在这个金额i的子集中

题解：

点击打开链接http://blog.csdn.net/tiantengtt/article/details/51885106

Pari wants to buy an expensive chocolate from Arya. She has n coins, the value of the i-th coin is c_i. The price of the chocolate is k, so Pari will take a subset of her coins with sum equal to k and give it to Arya.

Looking at her coins, a question came to her mind: after giving the coins to Arya, what values does Arya can make with them? She is jealous and she doesn't want Arya to make a lot of values. So she wants to know all the values x, such that Arya will be able to make xusing some subset of coins with the sum k.

Formally, Pari wants to know the values x such that there exists a subset of coins with the sum k such that some subset of this subset has the sum x, i.e. there is exists some way to pay for the chocolate, such that Arya will be able to make the sum x using these coins.

Input

The first line contains two integers n and k (1  ≤  n, k  ≤  500) — the number of coins and the price of the chocolate, respectively.

Next line will contain n integers c₁, c₂, ..., c_n (1 ≤ c_i ≤ 500) — the values of Pari's coins.

It's guaranteed that one can make value k using these coins.

Output

First line of the output must contain a single integer q— the number of suitable values x. Then print q integers in ascending order — the values that Arya can make for some subset of coins of Pari that pays for the chocolate.

Example

Input

6 185 6 1 10 12 2

Output

160 1 2 3 5 6 7 8 10 11 12 13 15 16 17 18 

Input

3 5025 25 50

Output

30 25 50 

#include<stdio.h>#include<string.h>#include<algorithm>#include<vector>#define maxn 505using namespace std;int main(){int n,k,len=0,dp[maxn][maxn]={0};scanf("%d%d",&n,&k);dp[0][0]=1;for(int i=0;i<n;i++){int a;scanf("%d",&a);for(int j=k;j>=a;j--){for(int l=0;l<=k-a;l++){if(dp[j-a][l]){dp[j][l]=1;dp[j][l+a]=1;}}}}for(int i=0;i<=k;i++){if(dp[k][i])len++;}printf("%d\n",len);for(int i=0;i<=k;i++){if(dp[k][i])printf("%d ",i);}return 0;}