#include <bits/stdc++.h>
#define int long long
using namespace std;
int n, k;
int a[305];
int sum[305];
// 前i个位置、放j个出口、第i个位置放/不放出口的最小逃生时间
int f[305][305][2];
int fLast[305][305][2]; // f[i][j][k] 对应的 last
int tLast[305][305][2]; // f[i][j][k] 对应的 last 位置是 0 还是 1
vector<int> ans;
signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin >> n >> k;
    for (int i = 1; i <= n; i++)
    {
        cin >> a[i];
        sum[i] = sum[i - 1] + a[i];
    }
    memset(f, 0x3f, sizeof(f));
    f[1][1][1] = 0;
    fLast[1][1][1] = -1;
    tLast[1][1][1] = -1;
    // 前面所有点到 i
    int sum = a[1];
    int cnt = a[1];
    for (int i = 2; i <= n; i++)
    {
        for (int j = 1; j <= min(i, k); j++)
        {
            // f[i][j][0]
            if (j < i)
            {
                int nowCnt = a[i];
                int nowSum = a[i];
                // a[last+1]~a[i] 都去 last 这个出口避难
                for (int last = i - 1; last >= j; last--)
                {
                    if (f[last][j][1] + nowSum <= f[i][j][0])
                    {
                        f[i][j][0] = f[last][j][1] + nowSum;
                        fLast[i][j][0] = last;
                        tLast[i][j][0] = 1;
                    }
                    nowCnt += a[last];
                    nowSum = nowSum + nowCnt;
                }
            }
            // f[i][j][1]
            if (j == 1)
            {
                f[i][j][1] = sum;
            }
            else
            {
                int nowSum = 0;
                // a[last+1]~a[i] 由 i 作为出口
                for (int last = i - 1; last >= j - 1; last--)
                {
                    if (f[last][j - 1][1] + nowSum <= f[i][j][1])
                    {
                        f[i][j][1] = f[last][j - 1][1] + nowSum;
                        fLast[i][j][1] = last;
                        tLast[i][j][1] = 1;
                    }
                    if (f[last][j - 1][0] + nowSum <= f[i][j][1])
                    {
                        f[i][j][1] = f[last][j - 1][0] + nowSum;
                        fLast[i][j][1] = last;
                        tLast[i][j][1] = 0;
                    }
                    nowSum += a[last] * (i - last);
                }
            }
        }
        cnt += a[i];
        sum += cnt;
    }
    int ii, jj, kk;
    ii = n, jj = k, kk = 1;
    if (f[n][k][0] <= f[n][k][1])
        kk = 0;
    while (jj != 0)
    {
        int ni, nj, nk;
        if (kk == 1)
        {
            ni = fLast[ii][jj][kk];
            nj = jj - 1;
            nk = tLast[ii][jj][kk];
            ans.push_back(ii);
        }
        else
        {
            ni = fLast[ii][jj][kk];
            nj = jj;
            nk = tLast[ii][jj][kk];
        }
        ii = ni, jj = nj, kk = nk;
    }
    cout << min(f[n][k][1], f[n][k][0]) << "\n";
    for (int i = ans.size() - 1; i >= 0; i--)
        cout << ans[i] << " ";
    cout << "\n";
    return 0;
}