动态规划—— E. The Values You Can Make

Pari wants to buy an expensive chocolate from Arya. She has n coins, the value of the i-th coin isci. The price of the chocolate isk, so Pari will take a subset of her coins with sum equal tok 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 valuesx, such that Arya will be able to make x using some subset of coins with the sumk.

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

Input

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

Next line will contain n integers c1, c2, ..., cn (1 ≤ ci ≤ 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 printq integers in ascending order — the values that Arya can make for some subset of coins of Pari that pays for the chocolate.

Examples

Input

6 18
5 6 1 10 12 2

Output

16
0 1 2 3 5 6 7 8 10 11 12 13 15 16 17 18 

Input

3 50
25 25 50

Output

3
0 25 50 

---

该题和背包比较相似，不过又加了个状态，就是是否可以组成s，则：
状态dp[i][j][p]为考虑到第i个数，当前所有数的和为j，组成和为p的子集是否可能。 
如果不使用第i个数，dp[i][j][p]=dp[i-1][j][p] 
如果使用第i个数但子集不取他，dp[i][j][p]=dp[i-1][j-ci][p] 
如果使用第i个数且子集取他，dp[i][j][p]=dp[i-1][j-ci][p-ci]

#include
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
typedef long long ll;
const int INF_INT  = 0x3f3f3f3f;
const ll INF_LL  = 1e18;
const int MAXN =1e5+5;
const int MAXM = 1015;


int n,k;
int c[505];
bool dp[505][505][505];

vector<int> ans;


int main()
{
//    ios_base::sync_with_stdio(false);
//    freopen("data.txt","r",stdin);
    while(cin>>n>>k)
    {
        dp[0][0][0]=1;
        for(int i=1;i<=n;i++)
        {
            cin >> c[i];
        }

        for(int z=1;z<=n;z++)
        {
            for(int i=0;i<=k;i++)
            {
                for(int j=0;j<=i&&j<=k;j++)
                {

                    if(dp[z-1][i][j] || i>=c[z]&&dp[z-1][i-c[z]][j] || j>=c[z]&&dp[z-1][i-c[z]][j-z[c]])
                        dp[z][i][j] = 1;
                }
            }
        }



        for(int i=0;i<=k;i++)
        {
            if(dp[n][k][i])
            {
                ans.push_back(i);
            }
        }
        cout<for(int i=0;icout<" ";
        }
        cout<