You are in a maze; seeing n doors in front of you in beginning. You can choose any door you like. The probability for choosing a door is equal for all doors. If you choose the i-th door, it can either take you back to the same position where you begun in xi minutes, or can take you out of the maze after xi minutes. If you come back to the same position, you can remember last K doors you have chosen. And when you are about to choose a door, you never choose a door that is already visited by you. Or we can say that you never choose a door that is visited as one of the last K doors. And the probability of choosing any remaining door is equal. Now you want to find the expected time to get out of the maze. Input Input starts with an integer T (≤ 100), denoting the number of test cases. EachcasecontainsablanklineandtwointegersnK(1≤n≤100,0≤K≤n). Thenextline contains n space separated integers. If the i-th integer (xi) is positive, you can assume that the i-th door will take you out of maze after xi minutes. If it’s negative, then the i-th door will take you back to the beginning position after abs(xi) minutes. You can safely assume that 1 ≤ abs(xi) ≤ 10000. Output For each case, print the case number and the expected time to get out of the maze. If it’s impossible to get out of the maze, print ‘-1.000’. Otherwise print the result rounded to three places after the decimal point. Add 10−9 to your result to avoid precision errors. Sample Input 4 20 10 10 20 10 -10 31 10 -10 -20 32 10 -10 -20 Sample Output Case 1: 10.000 Case 2: 20.000 Case 3: 30.000 Case 4: 25.000