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: Accepted)

#include <iostream>
#include <vector>
using namespace std;

int N,count = 0;
bool judge = true;
vector<vector<int>> v(100010);
int visited[100010] = {0},col[100010] = {0};


bool dfs(int n){
	visited[n] = 1;
	bool j = true;
	if(n!=1 && v[n].size()==1) return true;
	for(int i=0;i<v[n].size();i++){
		int s = v[n][i];
		if(visited[s]==0 && dfs(s)){
			if(col[n]==0){
				col[n] = 1;
				count++;
				j = false;
			}else judge = false;
		}
	}
	return j;
}


int main(){
	cin >> N;
	int a,b;
	for(int i=0;i<N-1;i++){
		cin >> a >> b;
		v[a].push_back(b);
		v[b].push_back(a);
	}
	if(N%2==1){
		cout << "First" << endl;
		return 0;
	}
	dfs(1);
/*	for(int i=1;i<=N;i++){
		cout << col[i] << (i!=N? " ":"\n");
	}
*/	cout << (!judge? "First":"Second") << endl;
}