Brain-Flak是一种图灵化的语言，在PPCG上享有很高的声誉。语言的记忆是由两个堆栈组成，而是一个“隐藏”的第三组被发现由瓦é在向导，导致大脑思维，高射炮方案的一些有趣的新方式。
那么，如何给那个可怜的隐藏第三堆栈更多的可见性呢？让我们创建一种语言，使第三个堆栈具有应有的识别力！在这里，我向您介绍Third-Flak。
语言
在Third-Flak中，只有一个堆栈，称为第三堆栈。运营商在他们的脑高射炮做同样的方式在第三堆栈上工作，但这里没有[]，{}，<>nilads没有{...}单子（所以在第三高射炮程序的唯一受理字符()[]<>）。这是每个运算符的工作（将给出示例，该示例表示第三个堆栈，并带有列表，其中最后一个元素是堆栈的顶部）：
()是Third-Flak中唯一的两个字符的运算符。它会将第三个堆栈的顶部加1。例如：[1,2,3]→[1,2,4](，[，<：未覆盖的前情况下，所有开口括号推0至第三层叠体。例如：[1,2,3]→[1,2,3,0])从第三个堆栈中弹出两个元素，并推回它们的总和。例如：[1,2,3]→[1,5]]从第三个堆栈中弹出两个元素，然后推回从第二个元素中减去第一个元素的结果。例如：[1,2,3]→[1,-1]>从第三个堆栈中弹出一个元素。示例[1,2,3]→[1,2]
这是该语言的其他规则：
在执行开始时，第三个堆栈仅包含单个0。
禁止在程序内部为空
[]或<>（如果遵循Third-Flak的语义，它们将始终是noop，但实际上它们在Brain-Flak中具有不同的含义，无法在此处重新创建）。括号总是需要平衡的，除非程序末尾的结尾的括号可能会丢失。例如，
[()<(()是一个有效的Third-Flak程序（程序末尾的第三个堆栈为[1,0,1]）。一个程序只能包含六个允许的字符
()[]<>。程序保证是非空的。
注意：以前的规则暗示，您不必处理需要从空堆栈中弹出的情况。
挑战
很简单，为Third-Flak写一个解释器。您的程序必须将Third-Flak程序作为输入，并在程序末尾返回第三堆栈的状态作为输出。
您的输出格式是灵活的，只要可以从中明确读取第三个堆栈的状态，并且始终以相同的方式编码相同的数字（这只是一种说法，任何输出格式都不是公然的方式）尝试作弊就可以了）。
您的输出选择可能会限制您可以管理的数字范围，只要这不会使挑战变得微不足道（因为这将是默认漏洞）。
测试用例
对于每个测试用例，第一行是输入，第二行是输出堆栈，用空格分隔的数字列表表示，其中堆栈的顶部是最后一个元素。
[()<(()
0 1 0 1
[((((()()()()()))
0 0 0 5
((([()][()][()])))
-3
[<<(((()()()())(((((
0 0 0 0 0 4 0 0 0 0 0
[()]<(([()])><[()]
-1 0 -1
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718 2
e是在这里。
[()]会违反我们不必担心从空堆栈中弹出的规则