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 <iostream>
#include <string>
#include <algorithm>
#include <vector>
#include <queue>
#include <map>
#include <set>
#include <cmath>
#include <numeric>
#include <iomanip>
#include <stack>
#include <complex>
#include <functional>
#include <tuple>

using namespace std;

#define Rep(i,a,b) for(ll i = a; i < b; ++i)
#define rep(i,b) Rep(i,0,b)
#define allof(a) (a).begin(), (a).end()

#define Yes(q) ((q) ? "Yes" : "No")
#define YES(q) ((q) ? "YES" : "NO")
#define Possible(q) ((q) ? "Possible" : "Impossible")
#define POSSIBLE(q) ((q) ? "POSSIBLE" : "IMPOSSIBLE")

using ll = long long;

constexpr int inf = 1e9 + 7;
constexpr ll infll = 1ll << 60ll;
constexpr ll mod = 1e9 + 7;
// 0~3までは右左下上 4~7までは斜め
constexpr int dx[] = { 1, 0, -1, 0, 1, 1, -1, -1 };
constexpr int dy[] = { 0, -1, 0, 1, 1, -1, -1, 1 };

namespace {
    template<typename T> bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
    template<typename T> bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }

    void Cout(long long x, const char* end = "\n") { std::cout << x << end; }
    template <typename T> void Cout(const T& x, const char* end = "\n") { std::cout << x << end; }
    template <typename T> void Cout(const std::vector<T>& x, const char* sep = " ", const char* end = "\n") { for (std::size_t i = 0, sz = x.size(); i < sz; ++i) { std::cout << x[i] << (i == sz - 1 ? end : sep); } }

    // 標準入出力
    struct inp {
        std::size_t szi, szj;
        inp(std::size_t _szi = 1, std::size_t _szj = 1) : szi(_szi), szj(_szj) {}
        template <typename T> operator T () const { T a; std::cin >> a; return a; }
        template <typename T> operator std::vector<T>() const { std::vector<T> a(szi); for (std::size_t i = 0; i < szi; ++i) std::cin >> a[i]; return a; }
        template <typename T> operator std::vector<std::vector<T>>() const {
            std::vector<std::vector<T>> a(szi, std::vector<T>(szj));
            for (std::size_t i = 0; i < szi; ++i) for (std::size_t j = 0; j < szj; ++j) cin >> a[i][j]; return a;
        }
    };
    inp inp1;

    struct Edge {
        ll from; // 移動前のノード
        ll to; // 移動先のノード
        ll cost; // 辺のコスト
        ll hoge; // 任意変更 いらないかも

        Edge() = default;
        Edge(ll _to) : from(-1), to(_to), cost(1), hoge(-1) {};
        Edge(ll _to, ll _cost) : from(-1), to(_to), cost(_cost), hoge(-1) {};
        Edge(ll _from, ll _to, ll _cost) : from(_from), to(_to), cost(_cost), hoge(-1) {};
        Edge(ll _from, ll _to, ll _cost, ll _hoge) : from(_from), to(_to), cost(_cost), hoge(_hoge) {};
        bool operator< (const Edge& rhs) const {
            return cost < rhs.cost;
        }
        bool operator> (const Edge& rhs) const {
            return cost > rhs.cost;
        }
    };

    using Graph = vector<vector<Edge>>;

    // ダイクストラ法 O(ElogV)
    vector<ll> dijkstra(ll start, const Graph& G) {
        vector<ll> dist(G.size(), infll);
        // 辺のコストが小さい順に取り出す　場合に応じて変更
        priority_queue<Edge, vector<Edge>, greater<Edge>> que;

        dist[start] = 0; // dist[i] := start->iまでの最短距離
        que.push(Edge(start, 0));

        while (!que.empty()) {
            ll cost, u; // 今までにかかった時間　現在の頂点
            cost = que.top().cost, u = que.top().to;
            que.pop();
            if (dist[u] < cost) continue;
            for (auto& e : G[u]) {
                ll v = e.to;
                ll new_cost = cost + e.cost;

                if (dist[v] <= new_cost) continue;

                dist[v] = new_cost;
                que.push(Edge(v, new_cost));
            }
        }
        return dist;
    }
}

int main() {

    int n = inp1;
    Graph G(n);
    vector<int> a(n - 1), b(n - 1);
    rep(i, n - 1) {
        cin >> a[i] >> b[i];
        --a[i], --b[i];
        G[a[i]].emplace_back(Edge(b[i]));
        G[b[i]].emplace_back(Edge(a[i]));
    }

    // 陣取りゲーム

    // お互いが近づこうとする
    auto dist_0 = dijkstra(0, G);
    auto dist_n = dijkstra(n - 1, G);

    int ans = 0;
    rep(i, n) {
        if (dist_0[i] <= dist_n[i]) ++ans;
        else --ans;
    }

    if (ans <= 0) Cout("Snuke");
    else Cout("Fennec");

    // データ構造の乱用

    return 0;
}