Problem:
Score : 900 points

Problem Statement

There is a tree with N vertices numbered 1 through N.
The i-th of the N-1 edges connects vertices a_i and b_i.

Initially, each vertex is uncolored.

Takahashi and Aoki is playing a game by painting the vertices. In this game, they alternately perform the following operation, starting from Takahashi:

Select a vertex that is not painted yet.

If it is Takahashi who is performing this operation, paint the vertex white; paint it black if it is Aoki.

Then, after all the vertices are colored, the following procedure takes place:

Repaint every white vertex that is adjacent to a black vertex, in black.

Note that all such white vertices are repainted simultaneously, not one at a time.

If there are still one or more white vertices remaining, Takahashi wins; if all the vertices are now black, Aoki wins.
Determine the winner of the game, assuming that both persons play optimally.

Constraints

2 ≤ N ≤ 10^5

1 ≤ a_i,b_i ≤ N

a_i ≠ b_i

The input 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

Print First if Takahashi wins; print Second if Aoki wins.

Sample Input 1

3
1 2
2 3

Sample Output 1

First

Below is a possible progress of the game:

First, Takahashi paint vertex 2 white.

Then, Aoki paint vertex 1 black.

Lastly, Takahashi paint vertex 3 white.

In this case, the colors of vertices 1, 2 and 3 after the final procedure are black, black and white, resulting in Takahashi's victory.

Sample Input 2

4
1 2
2 3
2 4

Sample Output 2

First

Sample Input 3

6
1 2
2 3
3 4
2 5
5 6

Sample Output 3

Second

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef vector<int> vi;
typedef pair<int,int> ii;

#define fill(a,x) memset(a,x,sizeof(a)) 
#define pb push_back
#define sz(x) (int)x.size() 
#define F first
#define S second
#define FOR(i,a,b) for(int i = a; i<=b; ++i)
#define NFOR(i,a,b) for(int i = a; i>=b; --i)
#define fast ios_base::sync_with_stdio(false),cin.tie(0),cout.tie(0)
const ll INF = 1e18;
const ll mod = 1e9+7;
const int N = 1e5+10; 

vi edges[N];
int deg[N];
bool yes[N];

int main(){
  fast;
  int n;
  cin>>n;
  fill(deg,0);
  fill(yes,0);

  FOR(i,1,n-1)
  {
  	int u,v;
  	cin>>u>>v;
  	edges[u].pb(v);
  	edges[v].pb(u);
  	deg[u]++;
  	deg[v]++;
  }

  FOR(i,1,n)
  {
  	int cnt=0;
  	FOR(j,0,sz(edges[i])-1)
  	{
  		if(deg[edges[i][j]]==1)
  			{
  				cnt++;
  				yes[i]=true;
  			}
  	}

  	if(cnt>1)
  	{
  		cout<<"First";
  		return 0;
  	}
  }

  FOR(i,1,n)
  {
  	int cnt=0;
  	FOR(j,0,sz(edges[i])-1)
  	{
  		if(yes[edges[i][j]])
  			cnt++;
  	}
  	if(cnt==sz(edges[i]))
  	{
  		cout<<"First";
  		return 0;
  	}
  }

  cout<<"Second";
  return 0;
}