/*
Idea:
- The order of cuts doesn't matter (only positions are important)
- dp[i][j] - best answer if we did j cuts and last cut was made after position i
- dp[i][j] = max (k < i)(dp[k][j - 1] + sum(k + 1, i) * (i + 1, n))
= max (k < i)(dp[k][j - 1] + (sum(1, i) - sum(1, k)) * ((1, n) - sum(1, i) ))
- If we take the right parts of formula above together, convex hull trick can be applied
- Because memory limit is tight, we can store only last 2 layers of dp
*/
#include <bits/stdc++.h>
using namespace std;
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define INF (1LL << 55)
#define MOD (1000 * 1000 * 1000 + 7)
#define maxn 100111
#define maxk 211
typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
ll k[maxn], n[maxn];
int ind[maxn];
int best_ind, top, bot;
ll dp[maxn][2], pre[maxn], arr[maxn];
int from[maxn][maxk];
bool bad(int l1, int l2, int l3){
return (n[l3] - n[l1]) * (k[l1] - k[l2]) <= (n[l2] - n[l1]) * (k[l1] - k[l3]);
}
void add(ll cur_k, ll cur_n, int cur_ind){
if(top - bot > 0 && k[top - 1] == cur_k){
if(cur_n > n[top - 1])
top--;
else
return;
}
k[top] = cur_k;
n[top] = cur_n;
ind[top] = cur_ind;
while(top - bot >= 3 && bad(top - 2, top - 1, top)){
k[top - 1] = k[top];
n[top - 1] = n[top];
ind[top - 1] = ind[top];
top--;
}
top++;
}
inline ll val(int ind, ll x){
return k[ind] * x + n[ind];
}
pair<ll, int> query(ll x){
if(best_ind >= top)
best_ind = top - 1;
while(best_ind + 1 < top && val(best_ind, x) <= val(best_ind + 1, x))
best_ind++;
return mp(val(best_ind, x), ind[best_ind]);
}
ll sum(int i, int j){
if(i > j)
return 0LL;
return pre[j] - pre[i - 1];
}
int main(){
int N, K;
scanf("%d%d", &N, &K);
for(int i = 1; i <= N; i++){
scanf("%lld", arr + i);
pre[i] = pre[i - 1] + arr[i];
}
for(int i = 0; i <= N; i++){
for(int j = 0; j < 2; j++)
dp[i][j] = -INF;
}
for(int i = 1; i <= N - 1; i++)
dp[i][1] = sum(1, i) * sum(i + 1, N);
for(int j = 2; j <= K; j++){
top = bot = 0;
best_ind = 0;
int jj = j % 2;
int prev = 1 - jj;
for(int i = j; i <= N - 1; i++){
add(sum(1, i - 1), dp[i - 1][prev] - sum(1, i - 1) * sum(1, N), i - 1);
pair<ll, int> cur = query(sum(1, i));
dp[i][jj] = cur.fi + sum(i + 1, N) * sum(1, i);
from[i][j] = cur.se;
}
}
int ans = 0;
for(int i = 1; i <= N - 1; i++){
if(dp[i][K % 2] > dp[ans][K % 2])
ans = i;
}
printf("%lld\n", dp[ans][K % 2]);
for(int i = K; i >= 1; i--){
printf("%d ", ans);
ans = from[ans][i];
}
return 0;
}
