레이팅 미적용In a parallel universe there are \(n\) chemical elements, numbered from 1 to \(n\). The element number \(n\) has not been discovered so far, and its discovery would be a pinnacle of research and would bring the person who does it eternal fame and the \(so-ca\)lled SWERC prize. There are \(m\) independent researchers, numbered from 1 to \(m\), that are trying to discover it. C\(ur- re\)ntly, the \(i-th\) researcher has a sample of the element \(s_{i}\). Every year, each researcher independently does one fusion experiment. In a fusion experiment, if the researcher currently has a sample of element \(a\), they produce a sample of an element \(b\) that is chosen uniformly at random between \(a+1\) and \(n\), and they lose the sample of element \(a\). The elements discovered by different researchers or in different years are completely independent. The first researcher to discover element \(n\) will get the SWERC prize. If several researchers discover the element in the same year, they all get the prize. For each \(i = 1\), 2, . . . , \(m\), you need to compute the probability that the \(i-th\) researcher wins the prize.
The first line contains two integers \(n\) and \(m\) (\(2 \le n \le 100\), \(1 \le m \le 10\)) — the number of elements and the number of researchers. The second line contains \(m\) integers \(s_{1}\), \(s_{2}\), . . . , \(s_{m}\) (\(1 \le s_{i} < n\)) — the elements that the researchers currently have.
Print \(m\) floati\(ng-po\)int numbers. The \(i-th\) number should be the probability that the \(i-th\) researcher wins the SWERC prize. Your answer is accepted if each number differs from the correct number by at most \(10^{−8}\).
2 3
1 1 1
1.0 1.0 1.0
3 3
1 1 2
0.5 0.5 1.0
3 3
1 1 1
0.625 0.625 0.625
100 7
1 2 4 8 16 32 64
0.178593469 0.179810455 0.182306771
0.187565366 0.199300430 0.229356322
0.348722518