Triples of Cows

There are initially $N-1$ pairs of friends among FJ's $N$ ($2\le N\le 2\cdot 10^5$) cows labeled $1\dots N$, forming a tree. The cows are leaving the farm for vacation one by one. On day $i$, the $i$th cow leaves the farm, and then all pairs of the $i$th cow's friends still present on the farm become friends.

For each $i$ from $1$ to $N$, just before the $i$th cow leaves, how many ordered triples of distinct cows $(a,b,c)$ are there such that none of $a,b,c$ are on vacation, $a$ is friends with $b$, and $b$ is friends with $c$?

The first line contains $N$.

The next $N-1$ lines contain two integers $u_i$ and $v_i$ denoting that cows $u_i$ and $v_i$ are initially friends ($1\le u_i,v_i\le N$).

• Inputs 4-5: $N\le 500$
• Inputs 6-10: $N\le 5000$
• Inputs 11-20: No additional constraints.

The answers for $i$ from $1$ to $N$ on separate lines.

$(1,2,3)$ and $(3,2,1)$ are the triples just before cow $1$ leaves.

After cow $1$ leaves, there are less than $3$ cows left, so no triples are possible.

At the beginning, cow $1$ is friends with all other cows, and no other pairs of cows are friends, so the triples are $(a, 1, c)$ where $a, c$ are different cows from $\{2, 3, 4\}$, which gives $3 \cdot 2 = 6$ triples.

After cow $1$ leaves, the remaining three cows are all friends, so the triples are just those three cows in any of the $3! = 6$ possible orders.

After cow $2$ leaves, there are less than $3$ cows left, so no triples are possible.