The Values You Can Make

Description（点击查看原题）

Pari wants to buy an expensive chocolate from Arya. She has n coins, the value of the i-th coin is ci. 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 x using 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, ..., 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 print q 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

题目分析

题意：一个人要用最多n个硬币凑出k元的价值，求用于凑成k元的硬币重新组合后，可以得到哪些价值，比如我们用价值为 1,4,5 的硬币凑成了k = 10 ，那么我们使用这三个硬币可以凑成价值有 0,1 ,4 ,5,6, 9,10 ，

解析：记dp[i][j] = 1 表示当前凑成了i 元，然后用凑成了 i 元的硬币可以凑成 j 元的价值，转移方程也很明显，dp[i][j] = max(dp[i-val][j] , dp[i-val][j-val]) ，其中我们将dp初始化为0，表示状态不存在，令dp[0][0] = 1 ，表示初始状态存在，因此状态转移方程也可以理解为只要dp[i-val][j] = 1 或者 dp[i-val][j-val] = 1 ，那么 dp[i][j] 就可以借助价值为val 的硬币从这两个状态转移而来了，所以dp[i][j] = 1。最后我们统计一下有多少 dp[k][j] = 1 并输出即可。

代码区

#include<iostream>
#include<cstdio>
#include<cstring>
//#define LOCAL = 1;
using namespace std;
typedef long long ll;int n, k;
int dp[510][510];   //dp[i][j] = 1 表示当前凑成了i 元，然后用凑成了 i 元的硬币可以凑成 j 元的价值
int val;int main()
{
#ifdef LOCALfreopen("input.txt", "r", stdin);freopen("output.txt", "w", stdout);
#endifwhile (scanf("%d%d", &n, &k) != EOF){memset(dp, 0, sizeof(dp));            //初始化为0，表示状态不存在dp[0][0] = 1;                        //最初的转移状态int tot = 0;                      //记录数量for (int i = 1; i <= n;i++){scanf("%d", &val);for(int j = k ; j >= val;j --){for(int t = k ; t>= 0 ; t --){if(dp[j-val][t] == 1)                 //前一个状态加上当前的硬币{if (dp[j][t] == 0 && j == k)     //统计数量，在第一次取得当前状态时统计tot++;dp[j][t] = 1;}if(t >= val && dp[j-val][t-val] == 1)   //前一个状态加上当前的硬币后，再用新硬币组成新的价值{if (dp[j][t] == 0 && j == k)     //统计数量，在第一次取得当前状态时统计tot++;dp[j][t] = 1;}}}}printf("%d\n", tot);for(int i = 0 ;i <= k ;i ++)  //输出满足的状态{if (dp[k][i] == 1)printf("%d ", i);}printf("\n");}return 0;
}

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codeforces 687C （动态规划）The Values You Can Make

The Values You Can Make

Description（点击查看原题）

Input

Output

Examples

Input

Output

Input

Output

题目分析

代码区

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