#include<bits/stdc++.h>
using namespace std;
const long long inf = (long long) 1e13 + 10;
const int inf1 = (int) 1e9 + 10;
#define int long long
#define ll long long
#define dbl long double
#define endl '\n'
#define sc second
#define fr first
#define mp make_pair
#define pb push_back
#define all(x) x.begin(), x.end()
#define maxn 100010
int n, k, a[maxn], ps[maxn];
int dp[maxn][2];
int32_t ant[maxn][202];
vector<pair<pair<int,int>,pair<pair<int,int>,int>>> cht;
//cht -> a,b,l,r
//chtlr -> l,r,a,b
int f(int a, int b, int x) {
return a*x + b;
}
dbl interx(int a1, int b1, int a2, int b2) {
return (dbl) (b2-b1)/(a1-a2);
}
void resetcht() {
cht.clear();
cht.pb(mp(mp(-inf1,+inf1),mp(mp(0,0),0)));
}
void attcht(int a, int b, int id) {
// so vai tirar para a esquerda
// vai tirando enquanto valer a pena
int ansl = inf;
while(cht.size()) {
int a1 = cht.back().sc.fr.fr;
int b1 = cht.back().sc.fr.sc;
int l = cht.back().fr.fr;
int r = cht.back().fr.sc;
int id1 = cht.back().sc.sc;
if(f(a,b,l) >= f(a1,b1,l)) {
cht.pop_back();
ansl = l;
}
else if(f(a,b,r) >= f(a1,b1,r)) {
int x = ceil(interx(a,b,a1,b1));
cht.pop_back();
cht.pb(mp(mp(l,x-1),mp(mp(a1,b1),id1)));
ansl = x;
break;
}
else {
break;
}
}
if(ansl <= inf1) {
cht.pb(mp(mp(ansl,inf1),mp(mp(a,b),id)));
}
}
int idq;
pair<int,int> qrr(int x) {
idq = min(idq,(int) cht.size()-1);
while(idq+1 != cht.size()) {
int a = cht[idq].sc.fr.fr;
int b = cht[idq].sc.fr.sc;
int a1 = cht[idq+1].sc.fr.fr;
int b1 = cht[idq+1].sc.fr.sc;
if(f(a1,b1,x) >= f(a,b,x)) idq++;
else break;
}
int a = cht[idq].sc.fr.fr;
int b = cht[idq].sc.fr.sc;
return mp(f(a,b,x),cht[idq].sc.sc);
}
void solve() {
cin >> n >> k;
for(int i = 1; i <= n; i++) {
cin >> a[i];
ps[i] = ps[i-1] + a[i];
}
for(int j = 1; j <= k; j++) {
resetcht();
//dp[0][j] = 0
idq = 0;
for(int i = j; i <= n; i++) {
auto qr = qrr(ps[i]);
dp[i][j&1] = qr.fr;
ant[i][j] = qr.sc;
attcht(ps[i],dp[i][j&1^1]-ps[i]*ps[i],i);
}
}
//dp[i][k] = max(dp[j-1][k-1] + ps[j-1]*(ps[i]-ps[j-1]))
//dp[j-1][k-1] - ps[j-1]² + ps[j-1]*ps[i]
//a ordem que vai colocando é crescente entao é só usar um vector bem mais simples
int id = n;
vector<int> ans;
for(int j = k; j >= 1; j--) {
ans.pb(ant[id][j]);
id = ant[id][j];
}
reverse(all(ans));
cout << dp[n][k&1] << endl;
for(auto x : ans) cout << x << " ";
cout << endl;
}
int32_t main() {
ios::sync_with_stdio(false); cin.tie(0);
// freopen("in.in", "r", stdin);
// freopen("out.out", "w", stdout);
int tt = 1;
// cin >> tt;
while(tt--) solve();
}
