#include <stdio.h>
#include <limits.h>

double f[3][20100] = {};

double min(double a, double b) {
	return a < b ? 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 = 1, old;
	for (i = 2; i <= n; i++) {
		for (j = 0; j <= p; j++) {
			if (j > 3 * i) break;

			now = i % 2 == 0 ? 2 : 1;
			old = now == 1 ? 2 : 1;

			f[now][j] = INT_MAX;
			if (j <= (i - 1) * 3) f[now][j] = (f[old][j] + arr[i].a);

			if (j - 1 >= 0 && j - 1 <= (i - 1) * 3) f[now][j] = min(f[now][j], (f[old][j - 1] + arr[i].b));
			if (j - 2 >= 0 && j - 2 <= (i - 1) * 3) f[now][j] = min(f[now][j], (f[old][j - 2] + arr[i].c));
			if (j - 3 >= 0 && j - 3 <= (i - 1) * 3) f[now][j] = min(f[now][j], (f[old][j - 3] + 1.0));

			/*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\n", f[now][p] / (double)n);

	return 0;
}