Codeforces Round #360 (Div. 2) -- E. The Values You Can Make （DP）

E. 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 sum k.

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 大体题意：给你n个数和k。问n个数所有能构成k的子集合中所有的可能的和是多少？思路：二维dp，  dp[i][j]表示当前和是i 能否构成j。如果dp[i][j]是可以的话，那么dp[i+m][j]和dp[i+m][j+m]都是可以得！（因为是子集合！！）最后枚举dp[K][i],把可以得放入ans数组或者vector输出即可！
#include<cstdio>#include<cstring>#include<algorithm>#include<vector>using namespace std;int dp[505][505];vector<int>ans;int main(){    int n,K;    while(scanf("%d %d",&n,&K) == 2){        ans.clear();        memset(dp,0,sizeof dp);        dp[0][0] = 1;        for (int i = 0; i < n; ++i){            int m;            scanf("%d",&m);            for (int j = K; j >= m; --j){                for (int k = 0; k + m <= K; ++k){                    if (dp[j-m][k])dp[j][k] = dp[j][k+m] = 1;                }            }        }        for (int i = 0; i <= K; ++i){            if (dp[K][i])ans.push_back(i);        }        sort(ans.begin(),ans.end());        int len = ans.size();        printf("%d\n",len);        for (int i = 0; i < len; ++i){            if (i)printf(" ");            printf("%d",ans[i]);        }        printf("\n");    }    return 0;}