#include <bits/stdc++.h>
using namespace std;
#define int long long
#define fi first
#define se second
#define pb push_back
#define freo(task) if(fopen(task".inp", "r")){ freopen(task".inp", "r", stdin); freopen(task".out", "w", stdout); }
#define FOR(i, a, b) for(int i = (a), _b = (b); i <= _b; i ++)
#define FORD(i, a, b) for(int i = (a), _b = (b); i >= _b; i --)
#define bit(x, i) ((x >> i) & 1)
#define oo 1e18
#define all(v) v.begin(), v.end()
using ll = long long;
using pii = pair<int, int>;
using vi = vector<int>;
// mt19937 rd(chrono::steady_clock::now().time_since_epoch().count());
// ll rand(ll l, ll r) { assert(l <= r); return uniform_int_distribution<ll>(l, r)(rd); }
const int N = 1e5 + 5;
const int mod = 1e9 + 7;
const int base = 31;
int n, k, a[N], dp[N];
/*
dp[i][j] là tổng lớn nhất khi chia i món cho j thằng
sub5: dp O(n ^ 3)
dp[i][j] = max(dp[i - 1][j], max(dp[k - 1][j]) + (a[k] -> i)
sub6: O(n ^ 2)
for(j 1->k)
dp_divide()
sub7: dp alien trick O(nlog(n))
*/
int dp1[305][305];
void sub5(){
FOR(i, 1, n) cin>>a[i], a[i] += a[i - 1];
FOR(i, 1, n)
FOR(j, 1, k){
dp1[i][j] = dp1[i - 1][j];
FOR(t, 1, i - 1)
dp1[i][j] = max(dp1[i][j], dp1[t - 1][j - 1] + a[i] - a[t - 1]);
}
// FOR(i, 1, n) cout<<dp1[i][k]<<" ";
cout<<dp1[n][k];
}
int dp2[2005][2005], g[2005][2005];
void sub6(){
FOR(i, 1, n) cin>>a[i];
FOR(i, 1, n)
FOR(j, 1, k){
dp2[i][j] = dp2[i - 1][j] + a[i];
dp2[i][j] = max(dp2[i][j], g[i - 1][j - 1] + a[i]);
g[i][j] = max(g[i - 1][j], dp2[i][j]);
}
int ans = 0;
FOR(j, 1, k) ans = max(ans, g[n][j]);
cout<<ans;
}
void solve(){
cin>>n>>k;
sub6();
}
int32_t main(){
ios::sync_with_stdio(false);
cin.tie(NULL);
// freo("");
int nTest = 1;
// cin>>nTest;
while(nTest --) solve();
// cerr << "\nTime: " << 1.0 * clock() / CLOCKS_PER_SEC << "s\n";
return 0;
}
/*
*/
