/*
Solution for APIO 2014 - Sequence
Tags : dp, Convex-hull Trick (CHT)
*/
#include <bits/stdc++.h>
using namespace std;
#pragma GCC target ("avx2")
#pragma GCC optimization ("Ofast")
#pragma GCC optimization ("unroll-loops")
typedef long long ll;
typedef vector<int> vi;
typedef pair<ll, ll> ii;
typedef vector<ii> vii;
typedef long double ld;
typedef pair<ll, ll> pll;
#define FOR(i, a, b) for(int i = a; i < b; i++)
#define ROF(i, a, b) for(int i = a; i >= b; i--)
#define ms memset
#define pb push_back
#define fi first
#define se second
ll MOD = 1000000007;
ll power(ll base, ll n){
if (n == 0) return 1;
if (n == 1) return base;
ll halfn = power(base, n/2);
if (n % 2 == 0) return (halfn * halfn) % MOD;
return (((halfn * halfn) % MOD) * base) % MOD;
}
ll inverse(ll n){
return power(n, MOD-2);
}
ll add(ll a, ll b){
return (a+b) % MOD;
}
ll mul(ll a, ll b){
a %= MOD;
return (a*b) % MOD;
}
ll gcd(ll a, ll b){
if (a == 1) return 1;
if (a == 0) return b;
return gcd(b%a, a);
}
const int N = 100001;
vector<ll> m(N), c(N);
ld g(ii l){
ll l1 = l.fi; ll l2 = l.se;
ld a = c[l1]-c[l2];
ld b = m[l2]-m[l1];
if (b==0) return -MOD;
return (ld) a/b;
}
void solve(){
// CHT to optimise O(n^2 * k) into O(n*k)
int n, k; cin >> n >> k;
vector<ll> a(n+1); FOR(i,1,n+1) cin >> a[i];
vector<ll> s(n+1); FOR(i,1,n+1) s[i]=a[i]+s[i-1];
ll dp[n+1][2]; // replace usual dp[n+1][k+1] to reduce memory usage for last subtask
int pos[n+1][k+1];
ms(dp,-1,sizeof(dp));
ms(pos,-1,sizeof(pos));
FOR(i,1,n+1){
dp[i][0]=s[i]*(s[n]-s[i]);
// only 1 component
}
FOR(j,1,k+1){
// m_l = 2*s[l]+s[n], m_l non-decreasing
// x = s[i]
// c_l = dp[l][j-1] - s[l] * (s[n]+s[l])
// f_l(x) = x^2 + m_l x + c_l
// l1 < l2, g(l1, l2) = (c_l1-c_l2)/(m_l2-m_l1)
// f_l1(x) <= f_l2(x) <==> g(l1, l2) <= x
// f_l1(x) >= f_l2(x) || f_l2(x) <= f_l3(x)
// g(l1, l2) >= x || g(l2, l3) <= x
// g(l1, l2) >= g(l2, l3) <==> l2 ignored
FOR(i,1,n+1){
dp[i][j%2]=-1; // dp values to compute later on
m[i] = 2*s[i]+s[n];
c[i] = dp[i][(j+1)%2] - s[i] * (s[n] + s[i]);
}
int best_l = 1;
// edge case : i = 2, only 1 possible value
ll val = dp[1][(j+1)%2] + (s[2]-s[1])*(s[n]-s[2]+s[1]);
if (val > dp[2][j%2] && dp[1][(j+1)%2] != -1){
dp[2][j%2]=val;
pos[2][j]=best_l;
}
list <ii> li; // stores pairs with increasing g(l1, l2) from left to right.
int r_l = 1;
FOR(i,3,n+1){
best_l = r_l;
ld cur;
if (dp[i-1][(j+1)%2] != -1){
r_l = i-1;
best_l = i-1;
ii c = {i-2, i-1};
cur = g(c);
while (!li.empty()){
ii u = li.back();
ld val = g(u);
if (val >= cur){
c.fi = u.fi;
cur = g(c);
// merge u and c since u.se, or c.fi can be ignored
}else{
break;
}
li.pop_back();
}
li.pb(c);
}
best_l = r_l;
cur = s[i];
while (!li.empty()){
if (cur < g(li.front())){
ii u = li.front();
best_l = u.fi;
break;
}
li.pop_front();
// remove values from the left since they are now smaller than cur
}
ll val1 = dp[best_l][(j+1)%2] + (s[i]-s[best_l])*(s[n]-s[i]+s[best_l]);
if (val1 > dp[i][j%2]){
dp[i][j%2]=val1;
//cout<<i<<" "<<j<<" "<<val<<" "<<val-dp[l][j-1]<<'\n';
pos[i][j]=best_l;
}
/*
FOR(l,1,i){
//O(K*N^2) brute force
ll val = dp[l][j-1] + (s[i]-s[l])*(s[n]-s[i]+s[l]);
if (val > dp[i][j] && dp[l][j-1] != -1){
dp[i][j]=val;
//cout<<i<<" "<<j<<" "<<val<<" "<<val-dp[l][j-1]<<'\n';
pos[i][j]=l;
}
}
*/
}
}
cout << dp[n][k%2]/2 << '\n';
int cur = n;
vi b;
ROF(i,k,1){
cur = pos[cur][i];
b.pb(cur);
}
ROF(i,k-1,0){
cout << b[i]<<" ";
}
cout << '\n';
return;
}
int main() {
ios::sync_with_stdio(false);
int TC = 1;
//cin >> TC;
FOR(i, 1, TC+1){
//cout << "Case #" << i << ": ";
solve();
}
}
