Problem:
Score : 800 points

Problem Statement

Takahashi and Aoki will play a game on a tree.
The tree has N vertices numbered 1 to N, and the i-th of the N-1 edges connects Vertex a_i and Vertex b_i.

At the beginning of the game, each vertex contains a coin.
Starting from Takahashi, he and Aoki will alternately perform the following operation:

Choose a vertex v that contains one or more coins, and remove all the coins from v.

Then, move each coin remaining on the tree to the vertex that is nearest to v among the adjacent vertices of the coin's current vertex.

The player who becomes unable to play, loses the game.
That is, the player who takes his turn when there is no coin remaining on the tree, loses the game.
Determine the winner of the game when both players play optimally.

Constraints

1 \leq N \leq 2 \times 10^5

1 \leq a_i, b_i \leq N

a_i \neq b_i

The graph given as input is a tree.

Input

Input is given from Standard Input in the following format:

N
a_1 b_1
a_2 b_2
:
a_{N-1} b_{N-1}

Output

Print First if Takahashi will win, and print Second if Aoki will win.

Sample Input 1

3
1 2
2 3

Sample Output 1

First

Here is one possible progress of the game:

Takahashi removes the coin from Vertex 1. Now, Vertex 1 and Vertex 2 contain one coin each.

Aoki removes the coin from Vertex 2. Now, Vertex 2 contains one coin.

Takahashi removes the coin from Vertex 2. Now, there is no coin remaining on the tree.

Aoki takes his turn when there is no coin on the tree and loses.

Sample Input 2

6
1 2
2 3
2 4
4 6
5 6

Sample Output 2

Second

Sample Input 3

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

Sample Output 3

First

Solution in C++14 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include "bits/stdc++.h"
using namespace std;
using ll = long long;
using vll = std::vector<ll>;
using vvll = std::vector<vll>;
using vvvll = std::vector<vvll>;
using dd = double;
using vdd = std::vector<dd>;
using vvdd = std::vector<vdd>;
using vvvdd = std::vector<vvdd>;

constexpr ll INF = 1LL << 60;
constexpr dd EPS = 1e-11;

//cin,cout高速化のおまじない＋桁数指定
struct Fast{
	Fast(){
		cin.tie(0);
		ios::sync_with_stdio(false);
		cout << fixed << setprecision(std::numeric_limits<double>::max_digits10);
	}
} fast;

#define REPS(i, S, E) for (ll i = (S); i <= (E); i++)
#define REP(i, N) REPS(i, 0, N-1)
#define DEPS(i, S, E) for (ll i = (E); i >= (S); i--)
#define DEP(i, N) DEPS(i, 0, N-1)
#define EACH(e, v) for (auto&& e : v)

template<class T> inline bool chmax(T &a, T b) { if (a < b) { a = b; return true; }return false; }
template<class T> inline bool chmin(T &a, T b) { if (a > b) { a = b; return true; }return false; }
template<class T> inline T MaxE(vector<T>&v,ll S,ll E){T m=v[S]; REPS(i,S,E)chmax(m,v[i]); return m;}//v[S]～v[E]の最大値
template<class T> inline T MinE(vector<T>&v,ll S,ll E){T m=v[S]; REPS(i,S,E)chmin(m,v[i]); return m;}//v[S]～v[E]の最小値
template<class T> inline T MaxE(vector<T> &v, ll N) { return MaxE(v, 0, N - 1); } //先頭N個中の最大値
template<class T> inline T MinE(vector<T> &v, ll N) { return MinE(v, 0, N - 1); }
template<class T> inline T MaxE(vector<T> &v) { return MaxE(v, (ll)v.size()); }
template<class T> inline T MinE(vector<T> &v) { return MinE(v, (ll)v.size()); }
template<class T> inline ll MaxI(vector<T>&v,ll S,ll E){ll m=S; REPS(i,S,E){if(v[i]>v[m])m=i;} return m;}
template<class T> inline ll MinI(vector<T>&v,ll S,ll E){ll m=S; REPS(i,S,E){if(v[i]<v[m])m=i;} return m;}
template<class T> inline ll MaxI(vector<T> &v, ll N) { return MaxI(v, 0, N - 1); }
template<class T> inline ll MinI(vector<T> &v, ll N) { return MinI(v, 0, N - 1); }
template<class T> inline ll MaxI(vector<T> &v) { return MaxI(v, (ll)v.size()); }
template<class T> inline ll MinI(vector<T> &v) { return MinI(v, (ll)v.size()); }
template<class T> inline T Sum(vector<T> &v, ll S, ll E){ T s=v[S]; REPS(i, S+1, E)s+=v[i]; return s; }
template<class T> inline T Sum(vector<T> &v, ll N) { return Sum(v, 0, N - 1); }
template<class T> inline T Sum(vector<T> &v) { return Sum(v, v.size()); }
template<class T> inline T POW(T a, ll n){ T r=1; for (; n>0; n>>=1, a*=a){ if (n&1)r*=a; } return r; }
inline ll POW(int a, ll n){ return POW((ll)a, n); }
inline ll MSB(ll a){for(ll o=63,x=-1;;){ll m=(o+x)/2; if(a<(1LL<<m))o=m; else x=m; if(o-x==1)return x;}}
inline ll CEIL(ll a, ll b){ return (a+b-1)/b; }

template<class T=ll> inline vector<T> cinv(ll N){ vector<T> v(N); REP(i, N)cin>>v[i]; return move(v);}
template<class T=ll, class S=ll> inline vector<pair<T, S>> cinv2(ll N){
	vector<pair<T, S>> v(N); REP(i,N){cin>>v[i].first>>v[i].second;} return move(v); }
template<class T=ll,class S=ll,class R=ll> inline vector<tuple<T, S, R>> cinv3(ll N){
	vector<tuple<T,S,R>> v(N); REP(i,N){cin>>get<0>(v[i])>>get<1>(v[i])>>get<2>(v[i]);} return move(v);}
template<class T=ll,class S=ll,class R=ll,class Q=ll> inline vector<tuple<T,S,R,Q>> cinv4(ll N){
	vector<tuple<T,S,R,Q>> v(N); REP(i,N){cin>>get<0>(v[i])>>get<1>(v[i])>>get<2>(v[i])>>get<3>(v[i]);} return move(v);}
template<class T> inline void coutv(vector<T> &v, char deli=' '){
	ll N=(ll)v.size(); REP(i,N){cout << v[i] << ((i==N-1)?'\n':deli);} }


struct mll{
	static ll MOD;
	ll val;
	mll(ll v = 0): val(v % MOD){ if (val < 0) val += MOD; }
	mll operator - () const { return -val; }
	mll operator + (const mll &b) const { return val + b.val; }
	mll operator - (const mll &b) const { return val - b.val; }
	mll operator * (const mll &b) const { return val * b.val; }
	mll operator / (const mll &b) const { return mll(*this) /= b; }
	mll operator + (ll b) const { return *this + mll(b); }
	mll operator - (ll b) const { return *this - mll(b); }
	mll operator * (ll b) const { return *this * mll(b); }
	friend mll operator + (ll a, const mll &b) { return b + a; }
	friend mll operator - (ll a, const mll &b) { return -b + a; }
	friend mll operator * (ll a, const mll &b) { return b * a; }
	mll &operator += (const mll &b) { val=(val+b.val)%MOD; return *this; }
	mll &operator -= (const mll &b) { val=(val+MOD-b.val)%MOD; return *this; }
	mll &operator *= (const mll &b) { val=(val*b.val)%MOD; return *this; }
	mll &operator /= (const mll &b) {
		ll c=b.val, d=MOD, u=1, v=0;
		while (d){
			ll t = c / d;
			c -= t * d; swap(c, d);
			u -= t * v; swap(u, v);
		}
		val = val * u % MOD;
		if (val < 0) val += MOD;
		return *this;
	}
	mll &operator += (ll b) { return *this += mll(b); }
	mll &operator -= (ll b) { return *this -= mll(b); }
	mll &operator *= (ll b) { return *this *= mll(b); }
	mll &operator /= (ll b) { return *this /= mll(b); }
	bool operator == (const mll &b) { return val == b.val; }
	bool operator != (const mll &b) { return val != b.val; }
	bool operator == (ll b) { return *this == mll(b); }
	bool operator != (ll b) { return *this != mll(b); }
	friend bool operator == (ll a, const mll &b) { return mll(a) == b.val; }
	friend bool operator != (ll a, const mll &b) { return mll(a) != b.val; }
	friend ostream &operator << (ostream &os, const mll &a) { return os << a.val; }
	friend istream &operator >> (istream &is, mll &a) { return is >> a.val; }
	static mll Combination(ll a, ll b){
		chmin(b, a-b);
		if (b<0) return mll(0);
		mll c = 1;
		REP(i, b) c *= a-i;
		REP(i, b) c /= i+1;
		return c;
	}
};
ll mll::MOD = 998244353LL;// (ll)(1e9 + 7);


vvll in_Graph(ll nodeNum, ll edgeNum, bool isDirected){
	vvll to(nodeNum);
	REP(i, edgeNum){
		ll v, u;
		cin >> v >> u;
		v--; u--;
		to[v].push_back(u);
		if (!isDirected) to[u].push_back(v);
	}
	return std::move(to);
}


ll Diameter(vvll &to){
	ll diameter = 0;

	std::function<ll(ll, ll)> dfsDiameter = [&](ll v, ll p)->ll{
		multiset<ll> depthSet;
		EACH(u, to[v]){
			if (u == p) continue;
			ll depth = dfsDiameter(u, v);
			depthSet.insert(dfsDiameter(u, v));
		}
		if (depthSet.empty()) return 0;
		if (depthSet.size() == 1){
			ll depth = *(depthSet.begin());
			chmax(diameter, depth+1);
			return depth+1;
		} else{
			auto it = depthSet.end();
			it--;
			ll maxDepth = *it;
			it--;
			ll secDepth = *it;
			chmax(diameter, maxDepth + secDepth + 2);
			return maxDepth+1;
		}
	};

	dfsDiameter(0, -1);
	return diameter;
}


void solve()
{
	ll N;
	cin >> N;
	auto to = in_Graph(N, N-1, false);

	ll dia = Diameter(to);

	//dfs(0, -1);
	//string ans = (diameter % 3 == 1) ? "Second" : "First";
	string ans = (dia % 3 == 1) ? "Second" : "First";
	cout << ans << '\n';
}

int main(){
	solve();
	return 0;
}