Codeforces Round #360 Editorial [+ Challenges!]

C. The Values You Can Make

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Pari wants to buy an expensive chocolate from Arya. She has n coins, the value of the i-th coin isc_i. 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, ..., 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 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

6 185 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

Hint

Use dynamic programming.

Solution

Let dp_i, j, k be true if and only if there exists a subset of the firsti coins with sumj, that has a subset with sumk. There are 3 cases to handle:

• The i-th coin is not used in the subsets.
• The i-th coin is used in the subset to makej, but it's not used in the subset of this subset.
• The i-th coin is used in both subsets.

So dp_i, j, k is equal todp_{i - 1, ji, k}OR dp_{i - 1, j - ci, k}OR dp_{i - 1, j - ci, k - ci}.

The complexity is O(nk²).

C++ code

//	   - -- --- ---- -----be name khoda----- ---- --- -- -		\\#include 
    
     using namespace std;inline int in() { int x; scanf("%d", &x); return x; }const int N = 505;bool dp[2][N][N];int main(){	int n, k;	cin >> n >> k;	dp[0][0][0] = 1;	for(int i = 1; i <= n; i++)	{		int now = i % 2;		int last = 1 - now;		int x = in();		for(int j = 0; j <= k; j++)			for(int y = 0; y <= j; y++)			{				dp[now][j][y] = dp[last][j][y];				if(j >= x)				{				    dp[now][j][y] |= dp[last][j - x][y];				    if(y >= x)    					dp[now][j][y] |= dp[last][j - x][y - x];				}			}	}	vector 
     
       res;	for(int i = 0; i <= k; i++)		if(dp[n % 2][k][i])			res.push_back(i);	cout << res.size() << endl;	for(int x : res)		cout << x << " ";	cout << endl;}
     
    

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