#include <iostream>
#include <vector>
#include <cstring>
#include <cmath>
using namespace std;
int n, m;
int minArea = 1 << 30;//记录最优解
int area = 0;//记录当前解
int minV[30], minA[30];//分别记录拼成i层最小可能的体积和表面积
int MaxVforNRH (int n, int r, int h) {
    int v;
    for (int i = 0; i < n; ++i) v += (r - i) * (r - i) * (h - i);
    return v;
}//计算在还有n层没拼完，当前一层的半径为r，高度为h时可能拼成的最大体积
void dfs (int v, int n, int r, int h) {
    if (n == 0) {
        if (v) return;
        else {
            minArea = min(minArea, area);
            return;
        }
    }
    if (v <= 0) return;//可行1
    if (minV[n] > v) return;//可行2
    if (area + minA[n] >= minArea) return;//最优
    if (h < n || r < n) return;//可行3
    if (MaxVforNRH(n, r, h) < v) return;//可行4
    for (int rr = r; rr >= n; --rr) {//爆搜
        if (n == m) area = rr * rr;
        for (int hh = h; hh >= n; hh--) {
            area += 2 * rr * hh;
            dfs (v - rr * rr * hh, n - 1, rr - 1, hh - 1);
            area -= 2 * rr * hh;
        }
    }
}
int main () {
    cin >> n >> m;
    minA[0] = 0;
    minV[0] = 0;
    for (int i = 1; i <= m; ++i) {
        minV[i] = minV[i - 1] + i * i * i;
        minA[i] = minA[i - 1] + 2 * i * i;
    }
    if (minV[m] > n) cout << 0 << endl;
    else {//处理搜索边界
        int maxH = (n - minV[m - 1]) / (m * m) + 1;
        int maxR = sqrt(double(n - minV[m - 1]) / m) + 1;
        area = 0;
        minArea = 1 << 30;
        dfs (n, m, maxR, maxH);
        if (minArea == 1 << 30) cout << 0 << endl;
        else cout << minArea << endl;
    }
    return 0;
}