#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define FOR(i,N) for(ll i = 0; i < N; i++)
#define all(x) (x).begin(), (x).end()
#define F first
#define S second
// APIO 14 Split The Sequence
/*
(0 indexed)
dp[k][m] = max score if we made k splits and the kth split was just BEFORE index m
dp[k][m] = max of: dp[k-1][i] + (ps[m]-p[i]) * (ps[N]-p[m]) for all (0 <= i < m)
nxt[k][m] = i where i was the optimal preceeding split point for split point m
0 indexed & splitting just BEFORE (code format) -> 1 indexed & splitting just AFTER (output format)
*/
const ll INF = 1e18;
const ll MAXN = 1e5 + 5;
const ll MAXK = 205;
ll N, K, PS[MAXN], dp[2][MAXN];
int nxt[MAXK][MAXN];
void solve(ll l, ll r, ll lOpt, ll rOpt, ll k) {
if (l > r) return ;
ll m = (l + r) / 2;
pair<ll,ll> bsf = {-INF, -1};
for(ll i = lOpt; i <= min(m-1, rOpt); i++) {
bsf = max(bsf, {dp[(k-1)%2][i] + (PS[m]-PS[i]) * (PS[N]-PS[m]), i});
}
dp[k%2][m] = bsf.F;
nxt[k][m] = bsf.S;
// cout << k << ' ' << m << ' ' << bsf.F << ' ' << bsf.S << ' ' << dp[k-1][bsf.S] << endl;
solve(l, m-1, lOpt, bsf.S, k);
solve(m+1, r, bsf.S, rOpt, k);
}
signed main() {
cin.tie(0); ios::sync_with_stdio(false);
cin >> N >> K;
FOR(i, N) {
cin >> PS[i+1];
PS[i+1] += PS[i];
}
for(ll k = 1; k <= K+1; k++) {
solve(1, N, 0, N-1, k);
}
cout << dp[(K+1)%2][N] << endl;
ll idx = N;
for(ll k = K+1; k > 1; k--) {
cout << nxt[k][idx] << ' ';
idx = nxt[k][idx];
}
}
