Codeforces 888E Maximum Subsequence

Discription

You are given an array a consisting of n integers, and additionally an integer m. You have to choose some sequence of indices b₁, b₂, ..., b_k (1 ≤ b₁ < b₂ < ... < b_k ≤ n) in such a way that the value of $Codeforces 888E Maximum Subsequence$ is maximized. Chosen sequence can be empty.

Print the maximum possible value of $Codeforces 888E Maximum Subsequence$ .

Input

The first line contains two integers n and m (1 ≤ n ≤ 35, 1 ≤ m ≤ 10⁹).

The second line contains n integers a₁, a₂, ..., a_n (1 ≤ a_i ≤ 10⁹).

Output

Print the maximum possible value of $Codeforces 888E Maximum Subsequence$ .

Example

Input

4 4
5 2 4 1

Output

Input

3 20
199 41 299

Output

Note

In the first example you can choose a sequence b = {1, 2}, so the sum $Codeforces 888E Maximum Subsequence$ is equal to 7 (and that's 3 after taking it modulo 4).

In the second example you can choose a sequence b = {3}.

(感觉看到n<=35的第一反应已经是折半搜索了hhh)

显然可以先搜出前n/2个数和后面剩的数可以得到的所有子集(包括空集)的结果(对m取模后)。

然后答案肯定就是从前面选一个子集，后面再选一个子集，把两个拼起来。

假设我们在前n/2中选的子集的权值是s,那么怎么在后面的数中找最优子集呢？

假设我们在后面的数中找的子集的权值是p，那么有两种情况：

1.s+p<m

2.m<=s+p<2*m

对于第一种情况，我们找到<m-s的最大的p即可；

对于第二种情况，我们找到最大的p即可(肯定得减掉一个m，所以p肯定是越大越好)

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#include<cmath>
#define ll long long
#define maxn 500005
using namespace std;
int a[maxn],n,m,b[maxn];
int num[105],tot,haf,ha;
int ans=0;

void dfs1(int x,int sum){
    if(sum>=ha) sum-=ha;
    if(x==haf){
        a[++n]=sum;
        return;
    }
    dfs1(x+1,sum);
    dfs1(x+1,sum+num[x+1]);
}

void dfs2(int x,int sum){
    if(sum>=ha) sum-=ha;
    if(x==tot){
        b[++m]=sum;
        return;
    }
    dfs2(x+1,sum);
    dfs2(x+1,sum+num[x+1]);
}

int main(){
    scanf("%d%d",&tot,&ha);
    for(int i=1;i<=tot;i++){
        scanf("%d",num+i);
        if(num[i]>=ha) num[i]-=num[i]/ha*ha;
    }
    haf=tot>>1;
    
    dfs1(0,0);
    dfs2(haf,0); 
    
    sort(a+1,a+n+1);
    sort(b+1,b+m+1);
    
    for(int i=1;i<=n;i++){
        int pos=lower_bound(b+1,b+m+1,ha-a[i])-b;
        ans=max(ans,a[i]+b[pos-1]);
        ans=max(ans,a[i]+b[m]-ha);
    }
    
    printf("%d\n",ans);
    return 0;
}