#include<bits/stdc++.h>
using namespace std;
#define watch(x) cout << (#x) << " is " << (x) << endl
#define fr(i,n) for(int i=0; i<n; i++)
#define rep(i, st, en) for(int i=st; i<=en; i++)
#define repn(i, st, en) for(int i=st; i>=en; i--)
#define sq(a) (a*a)
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
ll mod = 1e9+7;
/// given an array of n integers, you have to maximize reward by doing k splits. Each split gives a reward of product of sums of left and right bounded subarray
vector<vector<ll>> dp;
vector<ll> a;
vector<ll> pref;
ll recurse(int st, int n, int k, ll psum){
ll &z = dp[st][k];
if(~z) return z;
if(!k) return z = 0ll;
ll p1, p2;
z = 0ll;
for(int i=st+1; i <= n - k; ++i){
p1 = pref[i - 1] - psum;
ll temp = recurse(i, n, k - 1, p1 + psum);
p2 = pref[n - 1] - pref[i - 1];
z = max(z, temp + p1*p2);
}
return z;
}
void solve(){
int n, k;
cin>>n>>k;
a.resize(n);
pref.resize(n, 0);
ll sum = 0ll;
fr(i, n) {
cin>>a[i];
sum += a[i];
pref[i] = sum;
}
/// order of the splits doesn't matter, associativity property holds!
dp.resize(n, vector<ll>(k + 1, -1));
ll res = recurse(0, n, k, 0ll);
cout<<res;
}
int main(){
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
ll t = 1;
// cin>>t;
while(t--){
solve();
}
return 0;
}
