D. Bookshelves+Avito Code Challenge 2018+数位-优快云博客

本文介绍了一个关于书架分配的问题，需要将一定数量的书籍分配到多个书架上，每个书架的价值定义为该书架上所有书籍价格之和。目标是在满足特定条件的情况下最大化所有书架价值的按位与结果。

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Mr Keks is a typical white-collar in Byteland.

He has a bookshelf in his office with some books on it, each book has an integer positive price.

Mr Keks defines the value of a shelf as the sum of books prices on it.

Miraculously, Mr Keks was promoted and now he is moving into a new office.

He learned that in the new office he will have not a single bookshelf, but exactly $k$

bookshelves. He decided that the beauty of the

k

shelves is the bitwise AND of the values of all the shelves.

He also decided that he won't spend time on reordering the books, so he will place several first books on the first shelf, several next books on the next shelf and so on. Of course, he will place at least one book on each shelf. This way he will put all his books on $k$

shelves in such a way that the beauty of the shelves is as large as possible. Compute this maximum possible beauty.

Input

The first line contains two integers $n$

and

k

(

1 \leq k \leq n \leq 50

) — the number of books and the number of shelves in the new office.

The second line contains $n$

integers

a_{1}, a_{2}, \dots a_{n}

, (

0 < a_{i} < 2^{50}

) — the prices of the books in the order they stand on the old shelf.

Output

Print the maximum possible beauty of $k$

shelves in the new office.

Examples

Input

Copy

10 4
9 14 28 1 7 13 15 29 2 31

Output

Copy

Input

Copy

7 3
3 14 15 92 65 35 89

Output

Copy

Note

In the first example you can split the books as follows:

$(9 + 14 + 28 + 1 + 7) & (13 + 15) & (29 + 2) & (31) = 24.$

In the second example you can split the books as follows:

$(3 + 14 + 15 + 92) & (65) & (35 + 89) = 64.$

#define happy

#include<bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef long double ld;

typedef pair<int,int> pi;
typedef pair<ll,ll> pl;
typedef pair<ld,ld> pd;

typedef vector<int> vi;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<pi> vpi;
typedef vector<pl> vpl;


#define rep(i,a,b) for(int i=a;i<=b;i++)
#define per(i,a,b) for(int i=b-1;i>=a;i--)

#define all(a) (a).begin(),(a).end()
#define sz(x) (int)(x).size()
#define mp make_pair
#define pb push_back
#define f first
#define s second



const ll INF=1e18;
const int MOD=1e9+7;
const int N=50+5;

ll rd(){
    ll x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}

int n,k;
ll a[N],s[N];
bool dp[N][N];

bool check(ll w){
    memset(dp,0,sizeof(dp));
    dp[0][0]=1;
    rep(i,0,n-1)rep(j,0,k)if(dp[i][j])rep(l,i+1,n)
    if(((s[l]-s[i])&w)==w)dp[l][j+1]=1;
    return dp[n][k];
}


int main(){
#ifdef happy
    freopen("in.txt","r",stdin);
#endif
    n=rd(),k=rd();
    ll ans=0;
    rep(i,1,n)a[i]=rd(),s[i]=s[i-1]+a[i];
    per(i,0,60)if(check(ans|(1ll<<i)))ans|=(1ll<<i);
    printf("%lld\n",ans);
}