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 $c$?

SAMPLE INPUT:

3
1 2
2 3

SAMPLE OUTPUT:

2
0
0

$(1,2,3)$ 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.

SAMPLE INPUT:

4
1 2
1 3
1 4

SAMPLE OUTPUT:

6
6
0
0

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.

SAMPLE INPUT:

5
3 5
5 1
1 4
1 2

SAMPLE OUTPUT:

8
10
2
0
0

Problem credits: Aryansh Shrivastava, Benjamin Qi