Given integer $n$ and $k$ such that $0 < k \leq n$ and $n$ piles of stones, two players alternate turns. By one move it is allowed to choose any $k$ piles and remove exactly one stone from each. The player who has to move but cannot is the loser. in the normal version of the game and (s)he is the winner in the misère version. Cases $k=1$ and $k = n$ are trivial. For $k=2$ the game was solved for $n \leq 6$. For $n \leq 4$ the Sprague-Grundy function was efficiently computed (for both versions). For $n = 5,6$ a polynomial algorithm computing P-positions was obtained for the normal version. \newline Then, for the case $k = n-1$, a very simple explicit rule that determines the Smith remoteness function was found for the normal version of the game: the player who has to move keeps a pile with the minimum even number of stones; if all piles have odd number of stones then (s)he keeps a maximum one, while the $n-1$ remaining piles are reduced by one stone each in accordance with the rules of the game. \newline Computations show that the same rule works efficiently for the misère version too. The exceptions are sparse and are listed in Section 2. Denote a position by $x = (x_1, \dots, x_n)$. Due to symmetry, we can assume wlog that $x_1 \leq \ldots \leq x_n$. Our computations partition all exceptions into the following three families: $x_1$ is even, $x_1 = 1$, and $x_1 \geq 3$ is odd. In all three cases we suggest explicit formulas that cover all found exceptions, but this is not proven.