#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
#define watch(x) cout<<(#x)<<"="<<(x)<<'\n'
#define mset(d,val) memset(d,val,sizeof(d))
#define setp(x) cout<<fixed<<setprecision(x)
#define forn(i,a,b) for(int i=(a);i<(b);i++)
#define fore(i,a,b) for(int i=(a);i<=(b);i++)
#define pb push_back
#define F first
#define S second
#define pqueue priority_queue
#define fbo find_by_order
#define ook order_of_key
typedef long long ll;
typedef pair<ll,ll> ii;
typedef vector<ll> vi;
typedef vector<ii> vii;
typedef long double ld;
typedef tree<ll, null_type, less<ll>, rb_tree_tag, tree_order_statistics_node_update> pbds;
void amin(ll &a, ll b){ a=min(a,b); }
void amax(ll &a, ll b){ a=max(a,b); }
void YES(){cout<<"YES\n";} void NO(){cout<<"NO\n";}
void SD(int t=0){ cout<<"PASSED "<<t<<endl; }
const ll INF = ll(1e18);
const int MOD = 998244353;
struct Line {
ll m, b;
Line(ll _m, ll _b) : m(_m), b(_b) {}
inline ll eval(ll x) { return m * x + b; }
};
struct ConvexHull {
deque<Line> d;
inline void clear() { d.clear(); }
bool irrelevant(Line Z) {
if(int(d.size()) < 2) return false;
Line X = d[int(d.size())-2], Y = d[int(d.size())-1];
return (X.b - Z.b) * (Y.m - X.m) <= (X.b - Y.b) * (Z.m - X.m);
}
void addline(ll m, ll b) {
Line l = Line(m,b);
while(irrelevant(l)) d.pop_back();
d.push_back(l);
}
Line query(ll x) {
if(d.empty()) return Line{0,0};
while(int(d.size()) > 1 && (d[0].b - d[1].b <= x * (d[1].m - d[0].m))) d.pop_front();
return d.front();
}
};
const bool DEBUG = 0;
const int MAXN = 100005;
const int MAXK = 205;
int n,K;
ll s[MAXN], dp[MAXK][MAXN], nxt[MAXK][MAXN];
int main()
{
ios_base::sync_with_stdio(0); cin.tie(0);
cin>>n>>K;
forn(i,0,n){
cin>>s[i];
if(i>0) s[i]+=s[i-1];
}
ConvexHull cht;
fore(i,1,K)
{
cht.clear();
forn(j,0,n)
{
Line res = cht.query(s[j]);
nxt[i][j] = res.m;
dp[i][j] = res.eval(s[j]);
cht.addline(s[j], dp[i-1][j]-s[j]*s[j]);
}
}
vi ans;
int cur=n-1;
for(int i=K;i>=1;i--){
for(int j=cur-1;j>=0;j--){
if(s[j]==nxt[i][cur]){
ans.pb(j);
cur=j;
break;
}
}
}
cout<<dp[K][n-1]<<'\n';
for(ll x: ans) cout<<x+1<<" ";
cout<<'\n';
return 0;
}
