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 Java (OpenJDK 11.0.6) (judge verdict: Accepted)

import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
import java.util.PriorityQueue;
import java.util.Queue;
import java.util.Scanner;
import java.util.stream.IntStream;

/**
 * 解説通りに作成したソースコード
 */
public class Main {

	public static void main(String[] args) {
		try (Scanner scanner = new Scanner(System.in)) {
			int n = scanner.nextInt();
			@SuppressWarnings("unchecked")
			List<Path>[] pathLists = new List[n];
			IntStream.range(0, n).forEach(i -> pathLists[i] = new ArrayList<>());
			IntStream.range(0, n - 1).forEach(i -> {
				int from = scanner.nextInt() - 1, to = scanner.nextInt() - 1;
				pathLists[from].add(new Path(from, to, 1));
				pathLists[to].add(new Path(to, from, 1));
			});
			// ノード1からの距離の配列
			int[] distances1 = dijkstra(pathLists, 0, n);
			// ノードnからの距離の配列
			int[] distancesn = dijkstra(pathLists, n - 1, n);
			// 黒と白に塗られたマスの数
			int black = 0, white = 0;
			for (int i = 0; i < n; i++) {
				if (distances1[i] <= distancesn[i]) {
					black++;
				} else {
					white++;
				}
			}
			System.out.println((black > white) ? "Fennec" : "Snuke");
		}
	}

	/**
	 * @param pathLists ノード間の連結パス一覧の配列
	 * @param src       開始ノード
	 * @param n         ノードの数
	 * @return 開始ノードからの距離の配列
	 */
	private static int[] dijkstra(List<Path>[] pathLists, int src, int n) {
		int[] distances = new int[n];
		Arrays.fill(distances, Integer.MAX_VALUE);
		// 最短距離が確定したかどうか
		boolean[] fixed = new boolean[n];
		Arrays.fill(fixed, false);
		// 出発地点までの距離を0とする
		distances[src] = 0;
		Queue<Path> queue = new PriorityQueue<>();
		queue.add(new Path(src, src, 0));
		while (!queue.isEmpty()) {
			int now = queue.poll().to;
			if (!fixed[now]) {
				fixed[now] = true;
				for (Path path : pathLists[now]) {
					if (!fixed[path.to] && distances[path.to] > distances[now] + path.cost) {
						distances[path.to] = distances[now] + path.cost;
						queue.add(new Path(path.to, path.to, distances[path.to]));
					}
				}
			}
		}
		return distances;
	}

	/**
	 * ノード間の一つのパスを表すクラス
	 */
	private static class Path implements Comparable<Path> {
		/** 開始ノード */
		@SuppressWarnings("unused")
		int from;
		/** 終了ノード */
		int to;
		/** 移動コスト */
		int cost;

		Path(int from, int to, int cost) {
			super();
			this.from = from;
			this.to = to;
			this.cost = cost;
		}

		@Override
		public int compareTo(Path path) {
			return Integer.compare(cost, path.cost);
		}
	}
}