CF39E.What Has Dirichlet Got to Do with That?

You all know the Dirichlet principle, the point of which is that if n boxes have no less than n + 1 items, that leads to the existence of a box in which there are at least two items.

Having heard of that principle, but having not mastered the technique of logical thinking, 8 year olds Stas and Masha invented a game. There are a different boxes and b different items, and each turn a player can either add a new box or a new item. The player, after whose turn the number of ways of putting b items into a boxes becomes no less then a certain given number n, loses. All the boxes and items are considered to be different. Boxes may remain empty.

Who loses if both players play optimally and Stas's turn is first?

你们都了解鸽巢原理（抽屉原理），其核心思想是：若 $n$ 个盒子中放置了不少于 $n+1$ 个物品，则必然存在至少一个盒子中包含不少于两个物品。

在听说该原理后，尽管尚未掌握逻辑推理技巧，但 8 岁的斯塔斯（Stas）和玛莎（Masha）发明了一款游戏。游戏中初始有 $a$ 个互不相同的盒子和 $b$ 个互不相同的物品；每回合，玩家可以选择添加一个新盒子或一个新物品。在某位玩家操作之后，若将 $b$ 个物品放入 $a$ 个盒子中的方案数（所有盒子与物品均视为互不相同，且允许盒子为空）变得不少于某个给定的数 $n$，则该玩家判负。

若双方均采取最优策略，且由斯塔斯先行，请问谁会输？

The only input line has three integers a, b, n (1 ≤ a ≤ 10000, 1 ≤ b ≤ 30, 2 ≤ n ≤ 109) — the initial number of the boxes, the number of the items and the number which constrains the number of ways, respectively. Guaranteed that the initial number of ways is strictly less than n.

唯一的一行输入包含三个整数 $a$、$b$、$n$（$1 \le a \le 10000$，$1 \le b \le 30$，$2 \le n \le 10^9$），分别表示初始的盒子数量、物品数量以及限制方案数的数值。保证初始的方案数严格小于 $n$。

Output "Stas" if Masha wins. Output "Masha" if Stas wins. In case of a draw, output "Missing".

如果玛莎获胜，输出“Stas”；如果斯塔斯获胜，输出“Masha”；若平局，则输出“Missing”。

In the second example the initial number of ways is equal to 3125.

• If Stas increases the number of boxes, he will lose, as Masha may increase the number of boxes once more during her turn. After that any Stas's move will lead to defeat.
• But if Stas increases the number of items, then any Masha's move will be losing.

在第二个例子中，初始的方案数为 3125。

• 如果斯塔斯增加箱子的数量，他将输掉游戏，因为玛莎可以在她的回合中再次增加箱子的数量。此后，斯塔斯的任何操作都将导致失败。
• 但如果斯塔斯增加物品的数量，则玛莎的任何操作都将导致失败。