// Arup Guha
// 3/16/2017
// Solution to 2017 UCF HS Contest Problem: Rolling Burrito

import java.util.*;

public class burrito {

	// What the problem asks us to use.
	final public static double MYPI = 3.141592653589793;

	public static void main(String[] args) {

		Scanner stdin = new Scanner(System.in);
		int numCases = stdin.nextInt();

		// Process each case.
		for (int loop=1; loop<=numCases; loop++) {

			// Get input but store in doubles.
			double vol = stdin.nextInt();
			double r = stdin.nextInt();

			// Calculate length
			double len = vol/MYPI/r/r;

			// Get the dimensions of the wrap.
			int width = stdin.nextInt();
			int length = stdin.nextInt();

			// Try both orientations.
			if (fits(len, r, length, width) || fits(len, r, width, length))
				System.out.println("Burrito #"+loop+": Don't worry, the burrito fits!");
			else
				System.out.println("Burrito #"+loop+": Looks like a cold burrito today.");
		}
	}

	// Returns true iff this burrito fits on the foilLen x foilWidth sheet (orientation fixed).
	public static boolean fits(double burritoLen, double burritoRad, double foilLen, double foilWidth) {

		double cir = 2*MYPI*burritoRad;

		// Building in a bit of a tolerance...here the circumference is too big.
		if (cir > foilWidth + 1e-10) return false;

		// Return whether or not the wrap is long enough.
		return burritoLen + 2*burritoRad < foilLen + 1e-10;
	}
}