Problem:
Score : 400 points

Problem Statement

Fennec and Snuke are playing a board game.

On the board, there are N cells numbered 1 through N, and N-1 roads, each connecting two cells. Cell a_i is adjacent to Cell b_i through the i-th road. Every cell can be reached from every other cell by repeatedly traveling to an adjacent cell. In terms of graph theory, the graph formed by the cells and the roads is a tree.

Initially, Cell 1 is painted black, and Cell N is painted white. The other cells are not yet colored.
Fennec (who goes first) and Snuke (who goes second) alternately paint an uncolored cell.
More specifically, each player performs the following action in her/his turn:

Fennec: selects an uncolored cell that is adjacent to a black cell, and paints it black.

Snuke: selects an uncolored cell that is adjacent to a white cell, and paints it white.

A player loses when she/he cannot paint a cell. Determine the winner of the game when Fennec and Snuke play optimally.

Constraints

2 \leq N \leq 10^5

1 \leq a_i, b_i \leq N

The given graph is a tree.

Input

Input is given from Standard Input in the following format:

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

Output

If Fennec wins, print Fennec; if Snuke wins, print Snuke.

Sample Input 1

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

Sample Output 1

Fennec

For example, if Fennec first paints Cell 2 black, she will win regardless of Snuke's moves.

Sample Input 2

4
1 4
4 2
2 3

Sample Output 2

Snuke

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.HashSet;
import java.util.LinkedList;
import java.util.Scanner;

public class Main 
{
	static class Node
	{
		HashSet<Node> e = new HashSet<>();
		int i;
		int dis1 = -1;
		int dis2 = -1;
		boolean census = false;
		int blacksize = 1;
		
		public Node(int id)
		{
			i=id;
		}
	}
	
	static int cen(Node n)
	{
//		System.out.println("census " + n.i);
		n.census = true;
		int score = 1;
		for(Node e : n.e)
		{
			if(!e.census)
			{
				e.census = true;
				score += cen(e);
			}
		}
		return score;
	}

	public static void main(String[] args) 
	{
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();

		Node[] nd = new Node[n];
		
		for(int i = 0; i < n; i++)
		{
			nd[i] = new Node(i);
		}
		
		for(int i = 0; i < n-1; i++)
		{
			int a = sc.nextInt()-1;
			int b = sc.nextInt()-1;
			nd[a].e.add(nd[b]);
			nd[b].e.add(nd[a]);
		}
		
		Node s1 = nd[0];
		Node s2 = nd[n-1];
		LinkedList<Node> q = new LinkedList<Node>();
		q.add(s1);
		s1.dis1 = 0;
		while(!q.isEmpty())
		{
			Node pos = q.pollFirst();
			for(Node e : pos.e)
			{
				if(e.dis1 == -1)
				{
					e.dis1 = pos.dis1+1;
					q.add(e);
				}
			}
		}
		
		q = new LinkedList<Node>();
		q.add(s2);
		s2.dis2 = 0;
		while(!q.isEmpty())
		{
			Node pos = q.pollFirst();
			for(Node e : pos.e)
			{
				if(e.dis2 == -1)
				{
					e.dis2 = pos.dis2+1;
					q.add(e);
				}
			}
		}
		
		int path = nd[0].dis2;
		Node fb = null;
		for(int i = 0; i < n; i++)
		{ //find frontline black
			Node x = nd[i];
			if(x.dis1 + x.dis2 == path && x.dis1 == path/2)
			{
				fb = x;
				break;
			}
		}
		
		int blackscore = 1;
		fb.census = true;
//		System.out.println("frontline " + fb.i + ", dis " + fb.dis1 + " " + fb.dis2);
		for(Node e : fb.e)
		{
//			System.out.println(fb.i + " " + e.i);
			if(!(e.dis1 + e.dis2 == path && e.dis2 < fb.dis2))
			{
//				System.out.println("Open census " + e.i + ", dis " + e.dis1 + " " + e.dis2);
				blackscore += cen(e);
			}
		}

		int whitescore = n-blackscore;
//		System.out.println(blackscore + " " + whitescore);
		if(blackscore>whitescore)
			System.out.println("Fennec");
		else System.out.println("Snuke");
	}

}