import sys
input = sys.stdin.readline

# Consider adding epsilon to a single breaks pushup count
# Greedily, we would want to add it to the break that
# would add minimal stress

# The optimal distribution will then be when the derivative
# of all functions evaluated at their distribution is
# all equal

# let the derivative be d:
#       d = 2x_i / t_i -> x_i = d*t_i / 2
#       Sum of d*t_i / 2 = p -> d = 2p / sum of t_i
#       x_i = p*t_i / sum of t_i
#       Stress = sum of (x_i ^ 2 / t_i)
#              = sum of ((p*t_i / sum of t_i)^2 / t_i)
#              = (p / sum of t_i)^2 * sum of (t_i^2 / t_i)
#              = p^2 / sum of t_i

def solve():
    n, p = map(int, input().split())
    ans = p**2 / sum(map(int, input().split()))
    print(f"{ans:0.10f}")

t = int(input())
for ti in range(t):
    solve()