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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include "bits/stdc++.h"
#include <sys/timeb.h>
#include <fstream>

using namespace std;

#define repl(i,a,b) for(int i=(int)(a);i<(int)(b);i++)
#define rep(i,n) repl(i,0,n)
#define replrev(i,a,b) for(int i=(int)(b)-1;i>=(int)(a);i--)
#define reprev(i,n) replrev(i,0,n)
#define repi(itr,ds) for(auto itr=ds.begin();itr!=ds.end();itr++)
#define all(a) a.begin(),a.end()
#define mp make_pair
#define mt make_tuple
#define INF 2000000000
#define INFL 1000000000000000000LL
#define EPS (1e-10)
#define MOD 1000000007
#define PI 3.1415926536
#define RMAX 4294967295

typedef long long ll;
typedef pair<int, int> P;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector<bool> vb;
typedef vector<char> vc;
typedef vector<string> vs;
typedef vector<double> vd;
typedef vector<P> vP;
typedef vector<vector<int> > vvi;
typedef vector<vector<bool> > vvb;
typedef vector<vector<ll> > vvll;
typedef vector<vector<char> > vvc;
typedef vector<vector<string> > vvs;
typedef vector<vector<double> > vvd;
typedef vector<vector<P> > vvP;
typedef priority_queue<int, vector<int>, greater<int> > pqli;
typedef priority_queue<ll, vector<ll>, greater<ll> > pqlll;
typedef priority_queue<P, vector<P>, greater<P> > pqlP;
typedef pair<int, pair<int, int> > Edge;
typedef vector<Edge> vE;
typedef priority_queue<Edge, vector<Edge>, greater<Edge> > pqlE;

int B = 0, W = 1, E = 2;
int dist = 0;
int N;
vvi G;
vb used;
vi col;

bool dfs(int pos, int par, int depth) {
	if (pos == N - 1) {
		dist = depth;
		return true;
	}
	bool flag = false;
	rep(i, G[pos].size()) {
		if (G[pos][i] == par)continue;
		flag = flag || dfs(G[pos][i], pos, depth + 1);
	}
	if (flag){
		if (depth <= dist / 2) {
			col[pos] = B;
		}
		else {
			col[pos] = W;
		}
	}
	return flag;
}

int main() {
	cin >> N;
	G = vvi(N);
	int bnum = 0;
	used = vb(N, false);
	used[0] = used[N - 1] = true;

	col = vi(N, E);
	col[0] = B;
	col[N - 1] = W;

	rep(i, N - 1) {
		int a, b;
		cin >> a >> b;
		a--; b--;
		G[a].push_back(b);
		G[b].push_back(a);
	}

	dfs(0, -1, 0);

	queue<pair<int, int>> Q;
	rep(i, N) {
		if (col[i] != E) {
			Q.push(mp(i, col[i]));
		}
	}
	while (!Q.empty()) {
		int pos = Q.front().first;
		int c = Q.front().second;
		Q.pop();
		rep(i, G[pos].size()) {
			if (col[G[pos][i]] != E)continue;
			Q.push(mp(G[pos][i], c));
			col[G[pos][i]] = c;
		}
	}

	rep(i, N) {
		if (col[i] == B)bnum++;

		//cout << col[i] << " ";
	}


	if (bnum > N - bnum) {
		cout << "Fennec" << endl;
	}
	else {
		cout << "Snuke" << endl;
	}

	return 0;
}