Python solution based on : * Adjancency List * Breadth First Search * Combination for pairing with each other ungrouped

def journeyToMoon(n, astronaut):

matrix = [ [] for _ in range(n) ] 

Vertex = set()

for a in astronaut:
    _a , _b = a
    matrix[ _a ].append(_b)
    matrix[ _b ].append(_a)

    Vertex.add(_a)
    Vertex.add(_b)


rank = 0

visited = [ False ] * n

print(Vertex)

_G = []

for i in Vertex:

    if not visited[i]:
        G = []
        q = [ i ]
        while q:

            curr = q[0]
            q = q[1:]

            G.append(curr)

            visited[curr] = True

            for x in matrix[curr]:
                if not visited[x]:
                    visited[x] = True
                    q.append(x)

        _G.append(G)

# unorganize
n -= sum([ len(g) for g in _G ])

print(f"re {n}")


# print(" --> ",_G)

result = 0

for i in range(len(_G)):
    _ii = len(_G[i])

    for j in range(i + 1 ,len(_G)):
        result += _ii * len(_G[j])

    result += _ii * n

return result + (( n * (n-1) ) // 2)