<p>You are an ant tasked with adding <code>n</code> new rooms numbered <code>0</code> to <code>n-1</code> to your colony. You are given the expansion plan as a <strong>0-indexed</strong> integer array of length <code>n</code>, <code>prevRoom</code>, where <code>prevRoom[i]</code> indicates that you must build room <code>prevRoom[i]</code> before building room <code>i</code>, and these two rooms must be connected <strong>directly</strong>. Room <code>0</code> is already built, so <code>prevRoom[0] = -1</code>. The expansion&nbsp;plan is given such that once all the rooms are built, every room will be reachable from room <code>0</code>.</p>

<p>You can only build <strong>one room</strong> at a time, and you can travel freely between rooms you have <strong>already built</strong> only if they are <strong>connected</strong>.&nbsp;You can choose to build <strong>any room</strong> as long as its <strong>previous room</strong>&nbsp;is already built.</p>

<p>Return <em>the <strong>number of different orders</strong> you can build all the rooms in</em>. Since the answer may be large, return it <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/06/19/d1.JPG" style="width: 200px; height: 212px;" />
<pre>
<strong>Input:</strong> prevRoom = [-1,0,1]
<strong>Output:</strong> 1
<strong>Explanation:</strong>&nbsp;There is only one way to build the additional rooms: 0 &rarr; 1 &rarr; 2
</pre>

<p><strong class="example">Example 2:</strong></p>
<strong><img alt="" src="https://assets.leetcode.com/uploads/2021/06/19/d2.JPG" style="width: 200px; height: 239px;" /></strong>

<pre>
<strong>Input:</strong> prevRoom = [-1,0,0,1,2]
<strong>Output:</strong> 6
<strong>Explanation:
</strong>The 6 ways are:
0 &rarr; 1 &rarr; 3 &rarr; 2 &rarr; 4
0 &rarr; 2 &rarr; 4 &rarr; 1 &rarr; 3
0 &rarr; 1 &rarr; 2 &rarr; 3 &rarr; 4
0 &rarr; 1 &rarr; 2 &rarr; 4 &rarr; 3
0 &rarr; 2 &rarr; 1 &rarr; 3 &rarr; 4
0 &rarr; 2 &rarr; 1 &rarr; 4 &rarr; 3
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>n == prevRoom.length</code></li>
	<li><code>2 &lt;= n &lt;= 10<sup>5</sup></code></li>
	<li><code>prevRoom[0] == -1</code></li>
	<li><code>0 &lt;= prevRoom[i] &lt; n</code> for all <code>1 &lt;= i &lt; n</code></li>
	<li>Every room is reachable from room <code>0</code> once all the rooms are built.</li>
</ul>
class Solution(object):

    modulo = 10**9+7

    # def ncr(self,n, r):
    #     r = max(r, n-r)
    #     # numer = reduce((lambda x, y: (x * y)%self.modulo), list(range(n, n-r, -1)))
    #     # denom = reduce((lambda x, y: (x * y)%self.modulo), list(range(1, r+1)))
    #     numer = functools.reduce(operator.mul, range(n, n-r, -1), 1)
    #     denom = functools.reduce(operator.mul, range(1, r+1), 1)
    #     return numer // denom 

    class Node:
        def __init__(self,v):
            self.val = v
            self.count=0
            self.child = []

    def count(self,node):
        # def update(x):
        #     s-=x
        #     return s
        if not node.child:
            return 1,1
        subchild_nums, ways = [],[]
        for child in node.child:
            subchild_num, way = self.count(child)
            subchild_nums += [subchild_num]
            ways += [way]
        s=sum(subchild_nums)
        l = []
        for i in range(len(ways)):
            l += [((math.comb(s,subchild_nums[i])) * (ways[i]))%self.modulo]# for i in range(len(ways))]
            s-=subchild_nums[i]
        # print([(subchild_nums[i:],subchild_nums[i],ways[i]) for i in range(len(ways))])
        # print(l)
        return sum(subchild_nums)+1,reduce((lambda x, y: ((x%self.modulo) * (y%self.modulo))%self.modulo), l)%self.modulo

    def waysToBuildRooms(self, prevRoom):
        """
        :type prevRoom: List[int]
        :rtype: int
        """
        nodes = [self.Node(i) for i in range(len(prevRoom))]
        # print(math.comb)
        for i,r in enumerate(prevRoom):
            if i==0:
                continue
            nodes[r].child.append(nodes[i])
        return int(self.count(nodes[0])[1])
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