  Given integer $n$ and $k$ such that $0 < k \leq n$ and $n$ piles of stones,
two player 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. 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 the normal and mis\`ere versions). For $n = 5,6$
a polynomial algorithm computing P-positions was obtained. Here we consider the
case $2 \leq k = n-1$ and compute Smith's remoteness function, whose even
values define the P-positions. In fact, an optimal move is always defined by
the following simple rule: if all piles are odd, keep a largest one and reduce
all other; if there exist even piles, keep a smallest one of them and reduce
all other. Such strategy is optimal for both players, moreover, it allows to
win as fast as possible from an N-position and to resist as long as possible
from a P-position.
