// APIO 2014 Problem 2 - Split the Sequence
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
const long long inf = 1LL << 62;
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<long long> dp(N + 1, inf);
vector<vector<int> > pre(K + 1, vector<int>(N + 1, -1));
dp[0] = 0;
for(int i = 1; i <= K; ++i) {
vector<long long> ndp(N + 1, inf);
vector<int> st = { 0 };
for(int j = 1; j <= N; ++j) {
int l = 0, r = st.size();
while(r - l > 2) {
int lc = (l * 2 + r) / 3;
int rc = (l + r * 2) / 3;
long long la = dp[st[lc]] - 1LL * S[st[lc]] * S[j];
long long lb = dp[st[rc]] - 1LL * S[st[rc]] * S[j];
if(la > lb) l = lc;
else r = rc;
}
for(int k = l; k < r; ++k) {
long long x = dp[st[k]] - 1LL * S[st[k]] * S[j];
if(ndp[j] > x) {
ndp[j] = x;
pre[i][j] = st[k];
}
}
ndp[j] += 1LL * S[j] * S[j];
while(st.size() >= 2) {
int sz = st.size();
__int128_t lc = __int128_t(dp[j] - dp[st[sz - 1]]) * (S[j] - S[st[sz - 2]]);
__int128_t rc = __int128_t(dp[j] - dp[st[sz - 2]]) * (S[j] - S[st[sz - 1]]);
if(lc <= rc) st.pop_back();
else break;
}
st.push_back(j);
}
dp = ndp;
}
// step #4. print the answer
cout << 1LL * S[N] * S[N] - dp[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;
}
