Discription
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 kkbookshelves. He decided that the beauty of the kk 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 kk 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 nn and kk (1≤k≤n≤501≤k≤n≤50) — the number of books and the number of shelves in the new office.
The second line contains nn integers a1,a2,…ana1,a2,…an, (0<ai<2500<ai<250) — the prices of the books in the order they stand on the old shelf.
Output
Print the maximum possible beauty of kk shelves in the new office.
Examples
10 4
9 14 28 1 7 13 15 29 2 31
24
7 3
3 14 15 92 65 35 89
64
Note
In the first example you can split the books as follows:
In the second example you can split the books as follows:
#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int maxn=65;
ll ci[maxn],sum[maxn],ans;
bool can[maxn][maxn];
int n,k;
inline bool solve(){
memset(can,0,sizeof(can));
can[0][0]=1;
for(int i=1;i<=n;i++)
for(int j=0;j<i;j++) if(((sum[i]-sum[j])&ans)==ans){
for(int u=0;u<=j;u++) if(can[j][u]) can[i][u+1]=1;
}
return can[n][k];
}
int main(){
ci[0]=1;
for(int i=1;i<=60;i++) ci[i]=ci[i-1]+ci[i-1];
scanf("%d%d",&n,&k);
for(int i=1;i<=n;i++) cin>>sum[i],sum[i]+=sum[i-1];
for(int i=60;i>=0;i--){
ans|=ci[i];
if(!solve()) ans^=ci[i];
}
cout<<ans<<endl;
return 0;
}