CF1225G.To Make 1

There are $n$ positive integers written on the blackboard. Also, a positive number $k \geq 2$ is chosen, and none of the numbers on the blackboard are divisible by $k$. In one operation, you can choose any two integers $x$ and $y$, erase them and write one extra number $f(x + y)$, where $f(x)$ is equal to $x$ if $x$ is not divisible by $k$, otherwise $f(x) = f(x / k)$.

In the end, there will be a single number of the blackboard. Is it possible to make the final number equal to $1$? If so, restore any sequence of operations to do so.

黑板上写着 $n$ 个正整数。此外，选定一个正整数 $k \geq 2$，且黑板上的所有数均不被 $k$ 整除。每次操作中，你可以任选两个整数 $x$ 和 $y$，将它们擦除，并额外写下数字 $f(x + y)$，其中函数 $f(x)$ 定义如下：若 $x$ 不被 $k$ 整除，则 $f(x) = x$；否则（即 $x$ 被 $k$ 整除），$f(x) = f(x / k)$。

最终，黑板上将仅剩一个数。能否通过若干次操作使得该最终数恰好等于 $1$？若可以，请还原出任意一种可行的操作序列。

The first line contains two integers $n$ and $k$ — the initial number of integers on the blackboard, and the chosen number ($2 \leq n \leq 16$, $2 \leq k \leq 2000$).

The second line contains $n$ positive integers $a_1, \ldots, a_n$ initially written on the blackboard. It is guaranteed that none of the numbers $a_i$ is divisible by $k$, and the sum of all $a_i$ does not exceed $2000$.

第一行包含两个整数 $n$ 和 $k$ —— 黑板上初始的整数个数，以及选定的数（$2 \leq n \leq 16$，$2 \leq k \leq 2000$）。

第二行包含 $n$ 个正整数 $a_1, \ldots, a_n$，它们初始写在黑板上。保证每个数 $a_i$ 均不能被 $k$ 整除，且所有 $a_i$ 的总和不超过 $2000$。

If it is impossible to obtain $1$ as the final number, print "NO" in the only line.

Otherwise, print "YES" on the first line, followed by $n - 1$ lines describing operations. The $i$-th of these lines has to contain two integers $x_i$ and $y_i$ to be erased and replaced with $f(x_i + y_i)$ on the $i$-th operation. If there are several suitable ways, output any of them.

如果无法得到最终数字 $1$，则在唯一一行中输出 "NO"。

否则，在第一行输出 "YES"，随后输出 $n - 1$ 行，每行描述一次操作。其中第 $i$ 行需包含两个整数 $x_i$ 和 $y_i$，表示在第 $i$ 次操作中将它们擦除并替换为 $f(x_i + y_i)$。若存在多种可行方案，输出任意一种即可。

In the second sample case:

• $f(8 + 7) = f(15) = f(5) = 5$;
• $f(23 + 13) = f(36) = f(12) = f(4) = 4$;
• $f(5 + 4) = f(9) = f(3) = f(1) = 1$.

在第二个样例中：

• $f(8 + 7) = f(15) = f(5) = 5$；
• $f(23 + 13) = f(36) = f(12) = f(4) = 4$；
• $f(5 + 4) = f(9) = f(3) = f(1) = 1$。