#pragma GCC optimize("Ofast,unroll-loops")
#pragma GCC target("avx2,bmi,bmi2,popcnt,lzcnt")
#include <iostream>
#include <vector>
#include <algorithm>
#include <deque>
#include <set>
#include <string>
#include <numeric>
#include <cmath>
#include <cassert>
#include <array>
#include <numeric>
using namespace std;
using ll = long long;
const ll INF = 1e18;
struct Line
{
ll k, b;
int who;
Line(ll k = 0, ll b = 0, int who = 0) : k(k), b(b), who(who)
{}
ll operator()(ll x)
{
return k * x + b;
}
};
bool check(Line a, Line b, Line c)
{
if (b.k == c.k) return b.b >= c.b;
return (__int128)(a.b - c.b) * (b.k - a.k) <= (__int128)(a.b - b.b) * (c.k - a.k);
}
int main()
{
cin.tie(0)->sync_with_stdio(0);
int n, k;
cin >> n >> k;
++k;
vector<ll> a(n);
for (ll &i : a) cin >> i;
array<vector<ll>, 2> dp;
dp[0] = dp[1] = vector<ll>(n + 1, INF);
dp[0][0] = 0;
ll res = accumulate(a.begin(), a.end(), 0ll);
res = res * res;
partial_sum(a.begin(), a.end(), a.begin());
a.insert(a.begin(), 0);
vector<vector<int>> parent(n + 1, vector<int>(k, -1));
auto dac = [&dp, &a, &parent](auto self, int l, int r, int l2, int r2, int level)
{
if (l > r) return;
int m = (l + r) >> 1, ind = l2;
for (int i = l2; i <= min(r2, m - 1); ++i)
{
ll cur = dp[0][i] + (a[m] - a[i]) * (a[m] - a[i]);
if (cur < dp[1][m])
{
dp[1][m] = cur;
ind = i;
}
}
parent[m][level] = ind;
self(self, l, m - 1, l2, ind, level);
self(self, m + 1, r, ind, r, level);
};
for (int i = 0; i < k; ++i)
{
//dac(dac, i + 1, n, 0, n - 1, i);
deque<Line> dq;
dq.push_back({-2 * a[i], dp[0][i] + a[i] * a[i], i});
for (int j = i + 1; j <= n; ++j)
{
while (dq.size() >= 2 && dq[0](a[j]) > dq[1](a[j])) dq.pop_front();
dp[1][j] = dq.front()(a[j]) + a[j] * a[j];
parent[j][i] = dq.front().who;
//add
Line add{-2 * a[j], dp[0][j] + a[j] * a[j], j};
while (dq.size() >= 2 && check(dq[(int)dq.size() - 2], dq.back(), add)) dq.pop_back();
if (dq.back().k != add.k) dq.push_back(add);
}
//i -> i + 1
dp[0] = dp[1];
dp[1] = vector<ll>(n + 1, INF);
}
vector<int> path(k - 1);
for (int i = k - 1, v = n; i > 0; v = parent[v][i], --i) path[k - i - 1] = parent[v][i];
cout << (res - dp[0].back()) / 2 << '\n';
reverse(path.begin(), path.end());
for (int i : path) cout << i << ' ';
return 0;
}
/*dp[i] = min j < i
dp[j] + (p[i] - p[j])^2 = dp[j] + p[j]^2 - 2p[j]*p[i] + p[i]^2
kx + b*/
/*
4
1 3
4
0 2 3
*/
