#include <bits/stdc++.h>
#define pii pair<int, int>
#define f first
#define s second
#define pb push_back
#define mp make_pair
#define ll long long
#define ld long double
#define sz(v) int(v.size())
#define all(v) v.begin(), v.end()
using namespace std;
const int N = (int) 1e4 + 7;
const int M = (int) 2e6 + 7;
const int K = 202;
const ll LINF = (ll) 1e18;
const int INF = (int) 1e9 + 7;
const double EPS = (double) 1e-9;
ll dp[N][K], a[N];
int p[N][K];
int n, k;
ll sum(int l, int r) {
return a[r] - a[l - 1];
}
ld inter(int i, int x, int y) {
return (dp[y][i] - dp[x][i]) * 1.0 / (a[y] - a[x]);
}
int st[N];
int l, r;
ll f(int i, int j, ll x) {
return x * 1ll * a[i] + dp[i][j];
}
int get(int j, int x) {
int id = -1;
for (auto it : st) {
if (id == -1 || f(it, j, x) > f(id, j, x))
id = it;
}
return id;
}
int main() {
#define fn "balls"
#ifdef witch
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
#else
// freopen(fn".in", "r", stdin);
// freopen(fn".out", "w", stdout);
#endif
cin >> n >> k;
for (int i = 1; i <= n; i++) {
cin >> a[i];
a[i] += a[i - 1];
}
for (int i = 1; i <= n; i++) {
dp[i][1] = sum(1, i) * sum(i + 1, n);
}
for (int i = 2; i <= k; i++) {
l = 1, r = 0;
for (int j = i - 1; j <= n; j++) {
while (r - l + 1 > 1 && inter(i - 1, st[r], j) - inter(i - 1, st[r - 1], st[r]) > -EPS) {
--r;
}
while (r - l + 1 > 1 && f(st[l], i - 1, -sum(j + 1, n)) <= f(st[l + 1], i - 1, -sum(j + 1, n))) {
++l;
}
/*
y[t] = (a[j] - a[t]) * (a[n] - a[j]) + dp[t][i - 1]
a[j]*a[n]-a[j]*a[j]-a[t]*a[n]+a[t]*a[j]
a[t]*(-a[n]+a[j])+dp[t][i-1]+a[j]*(a[n]-a[j])
k=a[j]-a[t]
*/
if (l <= r) {
int t = st[l];
if (dp[j][i] < dp[t][i - 1] + sum(j + 1, n) * sum(t + 1, j)) {
dp[j][i] = dp[t][i - 1] + sum(j + 1, n) * sum(t + 1, j);
p[j][i] = t;
}
t = st[r];
if (dp[j][i] < dp[t][i - 1] + sum(j + 1, n) * sum(t + 1, j)) {
dp[j][i] = dp[t][i - 1] + sum(j + 1, n) * sum(t + 1, j);
p[j][i] = t;
}
}
st[++r] = j;
}
}
int j = 1;
for (int i = 1; i <= n; i++) {
if (dp[j][k] < dp[i][k]) j = i;
}
cout << dp[j][k] << endl;
vector<int> ans;
while (j > 0 && k > 0) {
ans.pb(j);
j = p[j][k];
k--;
}
reverse(all(ans));
for (auto it : ans) cout << it << " ";
return 0;
}
