CF1776K.Uniform Chemistry

In 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-called SWERC prize.

There are $m$ independent researchers, numbered from $1$ to $m$, that are trying to discover it. Currently, 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 $i = 1, \, 2, \, \dots, \, m$, you need to compute the probability that the $i$-th researcher wins the prize.

在一个平行宇宙中，存在 $n$ 种化学元素，编号从 $1$ 到 $n$。其中，编号为 $n$ 的元素尚未被发现；它的发现将标志着科研的巅峰成就，并使发现者获得永恒的声誉以及所谓的 SWERC 奖。

共有 $m$ 位彼此独立的研究员，编号从 $1$ 到 $m$，他们正致力于发现该元素。目前，第 $i$ 位研究员手中拥有一种元素 $s_i$ 的样品。每年，每位研究员各自独立地进行一次聚变实验：若某研究员当前持有元素 $a$ 的样品，则他在实验中会以等概率在区间 $[a+1, n]$ 中随机选择一个整数 $b$，并由此生成元素 $b$ 的样品，同时失去元素 $a$ 的样品。不同研究员之间、或不同年份中所发现的元素完全相互独立。

首位发现元素 $n$ 的研究员将独享 SWERC 奖；若多位研究员于同一年发现该元素，则他们共同获奖。对每个 $i = 1,\,2,\,\dots,\,m$，你需要计算第 $i$ 位研究员赢得该奖的概率。

The first line contains two integers $n$ and $m$ ($2 \le n \le 10^{18}$, $1 \le m \le 100$) — the number of elements and the number of researchers.

The second line contains $m$ integers $s_1, \, s_2, \, \dots, \, s_m$ ($1 \le s_i \lt n$) — the elements that the researchers currently have.

第一行包含两个整数 $n$ 和 $m$（$2 \le n \le 10^{18}$，$1 \le m \le 100$）—— 分别表示元素总数和研究人员数量。

第二行包含 $m$ 个整数 $s_1, \, s_2, \, \dots, \, s_m$（$1 \le s_i \lt n$）—— 表示研究人员当前拥有的元素。

Print $m$ floating-point 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 $10^{-8}$.

输出 $m$ 个浮点数。其中第 $i$ 个数应为第 $i$ 位研究人员赢得 SWERC 奖项的概率。若每个数与正确答案的差值均不超过 $10^{-8}$，则你的答案将被接受。

In the first sample, all researchers will discover element $2$ in the first year and win the SWERC prize.

In the second sample, the last researcher will definitely discover element $3$ in the first year and win the SWERC prize. The first two researchers have a $50\%$ chance of discovering element $2$ and a $50\%$ chance of discovering element $3$, and only element $3$ will bring them the prize.

In the third sample, each researcher has an independent $50\%$ chance of discovering element $3$ in the first year, in which case they definitely win the SWERC prize. Additionally, if they all discover element $2$ in the first year, which is a $12.5\%$ chance, then they will all discover element $3$ in the second year and all win the prize.

在第一个样例中，所有研究人员都将在第一年发现元素 $2$，并赢得 SWERC 奖项。

在第二个样例中，最后一名研究人员必定在第一年发现元素 $3$，并赢得 SWERC 奖项。前两名研究人员各有 $50\%$ 的概率发现元素 $2$、$50\%$ 的概率发现元素 $3$，而只有发现元素 $3$ 才能让他们获奖。

在第三个样例中，每名研究人员在第一年独立地以 $50\%$ 的概率发现元素 $3$，此时他们必定赢得 SWERC 奖项。此外，若他们全部在第一年发现元素 $2$（该事件发生的概率为 $12.5\%$），则他们将在第二年全部发现元素 $3$，并全部获奖。