Problem:
Score : 800 points

Problem Statement

Takahashi and Aoki will play a game on a tree.
The tree has N vertices numbered 1 to N, and the i-th of the N-1 edges connects Vertex a_i and Vertex b_i.

At the beginning of the game, each vertex contains a coin.
Starting from Takahashi, he and Aoki will alternately perform the following operation:

Choose a vertex v that contains one or more coins, and remove all the coins from v.

Then, move each coin remaining on the tree to the vertex that is nearest to v among the adjacent vertices of the coin's current vertex.

The player who becomes unable to play, loses the game.
That is, the player who takes his turn when there is no coin remaining on the tree, loses the game.
Determine the winner of the game when both players play optimally.

Constraints

1 \leq N \leq 2 \times 10^5

1 \leq a_i, b_i \leq N

a_i \neq b_i

The graph given as input is a tree.

Input

Input is given from Standard Input in the following format:

N
a_1 b_1
a_2 b_2
:
a_{N-1} b_{N-1}

Output

Print First if Takahashi will win, and print Second if Aoki will win.

Sample Input 1

3
1 2
2 3

Sample Output 1

First

Here is one possible progress of the game:

Takahashi removes the coin from Vertex 1. Now, Vertex 1 and Vertex 2 contain one coin each.

Aoki removes the coin from Vertex 2. Now, Vertex 2 contains one coin.

Takahashi removes the coin from Vertex 2. Now, there is no coin remaining on the tree.

Aoki takes his turn when there is no coin on the tree and loses.

Sample Input 2

6
1 2
2 3
2 4
4 6
5 6

Sample Output 2

Second

Sample Input 3

7
1 7
7 4
3 4
7 5
6 3
2 1

Sample Output 3

First

Solution in Python (3.4.3) (judge verdict: Runtime Error)

# この問題はグラフの最大経路を各ターンに1or2減らし、自分のターンに0にした方が負けというゲームであると言える
# 先攻のターンの時点で最後に枝数が1つなら後攻が、2つか3つなら先攻が勝つ→最大経路を3で割った余りで場合分け
# 後攻は先行の手に関係なく１ターンの枝数の減少を3にできる→初期が3n+1なら後攻が必勝
# 先攻は常に3n or 3n+2を維持すれば勝てる、これは3nの時は1を3n+2の時は2を出していれば可能
# つまり最大経路が3n or 3n+2なら先行、3n+1なら後攻が勝つ

n = int(input())

dimensionDict = {}
edgeList = []
leafList = []
for i in range(n):
  dimensionDict[(i+1)] = 0
for i in range(n-1):
  edge0, edge1 = map(int, input().split())
  dimensionDict[edge0] += 1
  dimensionDict[edge1] += 1
  edgeList.append((edge0, edge1))
  
for k, v in dimensionDict.items():
  if v == 1:
    leafList.append(k)

for k in leafList:
  distanceDict[k] = 0
  for edge in edgeList:
    if edge[0] == k:
      distanceDict[edge[1]] =  distanceDict[k] + 1
    elif edge[0] == k:
      distanceDict[edge[0]] =  distanceDict[k] + 1
# 途中です