// APIO 2014 Problem 2 - Split the Sequence
#include <vector>
#include <iostream>
#include <algorithm>
#pragma GCC target("avx")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
using namespace std;
const long long inf = 1LL << 60;
int main() {
// step #1. read input
cin.tie(0);
ios_base::sync_with_stdio(false);
int N, K;
cin >> N >> K; ++K;
vector<int> A(N);
for(int i = 0; i < N; ++i) {
cin >> A[i];
}
// step #2. preparation for calculating the answer
vector<int> S(N + 1);
for(int i = 0; i < N; ++i) {
S[i + 1] = S[i] + A[i];
}
// step #3. calculate the answer with dynamic programming
vector<vector<long long> > dp(K + 1, vector<long long>(N + 1, inf));
vector<vector<int> > pre(K + 1, vector<int>(N + 1, -1));
dp[0][0] = 0;
for(int i = 1; i <= K; ++i) {
for(int j = 1; j <= N; ++j) {
for(int k = 0; k < j; ++k) {
if(dp[i][j] > dp[i - 1][k] - 1LL * S[j] * S[k]) {
dp[i][j] = dp[i - 1][k] - 1LL * S[j] * S[k];
pre[i][j] = k;
}
}
dp[i][j] += 1LL * S[j] * S[j];
}
}
// step #4. print the answer
cout << 1LL * S[N] * S[N] - dp[K][N] << endl;
int pos = N;
for(int i = K; i >= 2; --i) {
cout << pre[i][pos] << (i != 2 ? ' ' : '\n');
pos = pre[i][pos];
}
return 0;
}
