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 C# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Collections.Generic;

public class Program
{
    private int N;
    private List<int>[] Edge;

    private int[] Dist;

    public void Solve()
    {
        var sc = new Scanner();
        N = sc.NextInt();
        Edge = new List<int>[N];
        for (int i = 0; i < N; i++)
        {
            Edge[i] = new List<int>();
        }

        for (int i = 0; i < N - 1; i++)
        {
            int a = sc.NextInt() - 1;
            int b = sc.NextInt() - 1;
            Edge[a].Add(b);
            Edge[b].Add(a);
        }

        // 双方の戦略として
        // 相手のところへ最短で行く
        // 自分で塗ったところで枝分かれしてるとこを塗っていく

        // 塗れるところが多いほうが勝ち

        Dist = new int[N];
        for (int i = 0; i < N; i++)
        {
            Dist[i] = int.MaxValue;
        }

        Dist[0] = 0;
        Search(0, -1);

        // 木の境目

        int pos = N - 1;
        int parent = -1;
        while (true)
        {
            foreach (var i in Edge[pos])
            {
                if (Dist[i] + 1 == Dist[pos])
                {
                    parent = pos;
                    pos = i;
                    break;
                }
            }

            if (Dist[pos] == Dist[N - 1] / 2)
            {
                break;
            }
        }

        //Console.WriteLine($"{pos} {parent}");

        int fennec = Count(pos, parent);

        int sunuke = N - fennec;
        Console.WriteLine(fennec > sunuke ? "Fennec" : "Snuke");
    }

    int Count(int i, int parent)
    {
        int ans = 1;
        foreach (int j in Edge[i])
        {
            if (j == parent) continue;
            ans += Count(j, i);
        }

        return ans;
    }

    void Search(int i, int parent)
    {
        foreach (int j in Edge[i])
        {
            if (j == parent) continue;
            Dist[j] = Dist[i] + 1;
            Search(j, i);
        }
    }

    public static void Main(string[] args)
    {
        new Program().Solve();
    }
}

class Scanner
{
    public Scanner()
    {
        _pos = 0;
        _line = new string[0];
    }

    const char Separator = ' ';
    private int _pos;
    private string[] _line;

    #region スペース区切りで取得

    public string Next()
    {
        if (_pos >= _line.Length)
        {
            _line = Console.ReadLine().Split(Separator);
            _pos = 0;
        }

        return _line[_pos++];
    }

    public int NextInt()
    {
        return int.Parse(Next());
    }

    public long NextLong()
    {
        return long.Parse(Next());
    }

    public double NextDouble()
    {
        return double.Parse(Next());
    }

    #endregion

    #region 型変換

    private int[] ToIntArray(string[] array)
    {
        var result = new int[array.Length];
        for (int i = 0; i < array.Length; i++)
        {
            result[i] = int.Parse(array[i]);
        }

        return result;
    }

    private long[] ToLongArray(string[] array)
    {
        var result = new long[array.Length];
        for (int i = 0; i < array.Length; i++)
        {
            result[i] = long.Parse(array[i]);
        }

        return result;
    }

    private double[] ToDoubleArray(string[] array)
    {
        var result = new double[array.Length];
        for (int i = 0; i < array.Length; i++)
        {
            result[i] = double.Parse(array[i]);
        }

        return result;
    }

    #endregion

    #region 配列取得

    public string[] Array()
    {
        if (_pos >= _line.Length)
            _line = Console.ReadLine().Split(Separator);

        _pos = _line.Length;
        return _line;
    }

    public int[] IntArray()
    {
        return ToIntArray(Array());
    }

    public long[] LongArray()
    {
        return ToLongArray(Array());
    }

    public double[] DoubleArray()
    {
        return ToDoubleArray(Array());
    }

    #endregion
}