// https://www.acmicpc.net/problem/10067
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
namespace CHT
{
typedef __int128 dll;
struct Line { ll a, b; int p; };
struct Frac
{
ll u, d;
Frac(ll _u, ll _d) : u(_u), d(_d) { if(d<0) u=-u, d=-d; }
bool operator < (const Frac &ot) const { return (dll)u*ot.d < (dll)ot.u*d; }
bool operator > (const Frac &ot) const { return (dll)u*ot.d > (dll)ot.u*d; }
};
ll divfloor(ll u, ll d) { return u/d - ((u^d)<0 && u%d); }
ll divceil(ll u, ll d) { return u/d + ((u^d)>0 && u%d); }
// min/max : ?, slope : ?, dir : ?
struct CHT
{
// Get cross point of line p, q
// If all queries are integer, change 'Frac' to 'div'
Frac cross(const Line &p, const Line &q) { return Frac(p.b-q.b, q.a-p.a); } // dir + : divfloor, dir - : divceil
deque<Line> V;
// Insert line p = ax+b
// b must be increasing (or decreasing) ('slope')
// cross(V[i-1], V[i]) < (or >) cross(V[i], V[i+1]) ('dir')
void push(Line p)
{
if(!V.empty() && V.back().a==p.a)
{
if(V.back().b <= p.b) return; // min : <= , max : >=
V.pop_back();
}
while(V.size()>1 && !(cross(V[V.size()-2], V[V.size()-1]) < cross(V[V.size()-1], p))) V.pop_back(); // dir + : <, dir - : >
V.push_back(p);
}
// Get min (or max) value at x in O(logN)
int query(ll x)
{
assert(!V.empty());
int lo=0, hi=V.size();
while(lo+1<hi)
{
int mid=lo+hi>>1;
if(cross(V[mid-1], V[mid]) < Frac(x, 1)) lo=mid; // dir + : <, dir - : >
else hi=mid;
}
// return V[lo].a*x+V[lo].b;
return V[lo].p;
}
// Get min (or max) value at x in ammortized O(1)
// x must be increasing (or decreasing) ('dir')
int query2(ll x)
{
assert(!V.empty());
while(V.size()>1 && cross(V[0], V[1]) < Frac(x, 1)) V.pop_front(); // dir + : <, dir - : >
// return V[0].a*x+V[0].b;
return V[0].p;
}
};
}
const int MAXN = 1e5;
int N, K;
ll A[MAXN+10];
namespace Alien
{
// MAXN must be defined
const int MAXN = 1e5;
int N;
// dp[i] : optimal dp value of i
// cnt[i] : how much transitions were used in optimal dp[i]
// memo[i] : previous transition position from i
// V : restored optimal solution
ll dp[MAXN+10];
int cnt[MAXN+10], memo[MAXN+10];
vector<int> V;
void init(int _N)
{
N=_N;
}
// For given lambda, calculate dp, cnt, memo, V
// dp[i] = min(or max)_{j<i} (dp[j] + cost(j, i)*2 - lambda)
void solve(ll lambda)
{
// initialize dp, cnt, memo, V, (and other data structures)
for(int i=0; i<=N; i++) dp[i]=cnt[i]=memo[i]=0;
V.clear();
CHT::CHT cht;
cht.push({0, 0, 0});
for(int i=1; i<=N; i++)
{
// get_opt(i), cost(p, q) must be implemented
// opt = argmin(or max)_{j<i} (dp[j] + cost(j, i)*2)
int opt = cht.query2(A[i]);
dp[i] = dp[opt] + (A[i]-A[opt])*(A[i]-A[opt])*2 - lambda; // Don't forget *2
cnt[i] = cnt[opt]+1;
memo[i] = opt;
cht.push({-4*A[i], 2*A[i]*A[i]+dp[i], i});
}
for(int i=N; i>0;)
{
V.push_back(i);
i=memo[i];
}
V.push_back(0);
reverse(V.begin(), V.end());
}
// Get optimal dp[N][K] for given K
// Returns (answer, restored solution)
// dp[i][k] = min(or max)_{j<i} (dp[j][k-1] + cost(j, i))
pair<ll, vector<int>> alien(int K)
{
// lambda equals slope
// minimum : K increase, lambda increase
// maximum : k increase, lambda decrease
ll lo=-1e18, hi=1e18; // range for lambda is [2*lo+1, 2*hi+1]
while(lo+1<hi)
{
ll mid=lo+hi>>1;
solve(2*mid+1);
if(K <= cnt[N]) hi=mid; // min : <= , max : >=
else lo=mid;
}
vector<int> P1, P2, ansV;
solve(2*lo+1); P1=V;
solve(2*hi+1); P2=V;
if(P1.size()>P2.size()) swap(P1, P2);
if(P1.size()-1==K) ansV=P1;
else if(P2.size()-1==K) ansV=P2;
else
{
assert(P1.size()-1<K && K<P2.size()-1);
int x=K+1-P1.size();
for(int i=0; i+1<P1.size() && i+x+1<P2.size(); i++)
{
if(P1[i]<=P2[i+x] && P2[i+x+1]<=P1[i+1])
{
for(int j=0; j<=i+x; j++) ansV.push_back(P2[j]);
for(int j=i+1; j<P1.size(); j++) ansV.push_back(P1[j]);
}
}
}
assert(ansV.size()==K+1);
solve(2*hi);
ll ans=dp[N]/2+hi*K;
return {ans, ansV};
}
}
int main()
{
scanf("%d%d", &N, &K); K++;
for(int i=1; i<=N; i++) scanf("%lld", &A[i]);
for(int i=1; i<=N; i++) A[i]+=A[i-1];
Alien::init(N);
auto [ans, ansV] = Alien::alien(K);
ans=(A[N]*A[N]-ans)/2;
printf("%lld\n", ans);
for(int i=1; i+1<ansV.size(); i++) printf("%d ", ansV[i]);
printf("\n");
}
