#include double f[3][15010] = {}; double min(double a, double i, double b, double j) { return a / i < b / j ? a : b; } int main() { int n, p; int i, j; struct { int a, b, c; } arr[5010]; scanf("%d%d", &n, &p); for (i = 1; i <= n; i++) { scanf("%d%d%d", &arr[i].a, &arr[i].b, &arr[i].c); } f[1][0] = arr[1].a; f[1][1] = arr[1].b; f[1][2] = arr[1].c; f[1][3] = 1.0; int now, old; for (i = 2; i <= n; i++) { for (j = 0; j <= p; j++) { if (j > 3 * i) continue; now = i % 2 == 0 ? 2 : 1; old = now == 1 ? 2 : 1; if (j <= (i - 1) * 3) { f[now][j] = (f[old][j] + arr[i].a); if (j - 1 >= 0) f[now][j] = min(f[now][j], i, (f[old][j - 1] + arr[i].b), i); if (j - 2 >= 0) f[now][j] = min(f[now][j], i, (f[old][j - 2] + arr[i].c), i); if (j - 3 >= 0) f[now][j] = min(f[now][j], i, (f[old][j - 3] + 1.0), i); } else { int d = j - (i - 1) * 3; switch (d) { case 1: f[now][j] = f[old][j - 1] + arr[i].b; break; case 2: f[now][j] = f[old][j - 2] + arr[i].c; break; case 3: f[now][j] = f[old][j - 3] + 1.0; break; } } } } printf("%.1lf", f[now][p] / (double)n); return 0; }