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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include <iostream>
#include <vector>
using namespace std;
const int MAXn = 1e5 + 10;
vector<int> g[MAXn];
int n, odd[MAXn], even[MAXn];

void dfs(int v, int par) {
	if (g[v].size() == 1 && par != 0)
		even[v] = 1;
	for (int i = 0; i < g[v].size(); i++)
		if (g[v][i] != par) {
			dfs(g[v][i], v);
			odd[v] += (even[g[v][i]]);
			even[v] += (odd[g[v][i]]);
		}
}

void dfs2(int v, int par) {
	if (!par)
		even[v]++;
	for (int i = 0; i < g[v].size(); i++)
		if (g[v][i] != par) {
			if (even[v] - (odd[g[v][i]]) > 0) {
				odd[g[v][i]]+= even[v] - (odd[g[v][i]]);
			}
			if (odd[v] - (even[g[v][i]]) > 0)
				even[g[v][i]]+= odd[v] - (even[g[v][i]]);
			dfs2(g[v][i], v);
		}	
	
}
int main() {
	cin >> n;
	for (int i = 0; i < n - 1; i++) {
		int u, v;
		cin >> u>> v;
		g[u].push_back(v);
		g[v].push_back(u);
	}
	dfs(1, 0);
	dfs2(1, 0);
//	for (int i = 1; i<= n; i++)
//		leaf += g[i].size() == 1;
	for (int i = 1; i <= n; i++) {
		int cnt = 0;
		for (int j = 0; j < g[i].size(); j++)
			if (g[g[i][j]].size() == 1)
				cnt++;
		if (cnt > 1)
			return cout << "First",0;
	}
	for (int i = 1; i <= n; i++) {
	//cout << i  <<' ' << g[i].size() << ' ' << even[i] << ' ' << odd[i] << endl;
		if ((!even[i] || !even[i]) && g[i].size() > 1) 
			return cout << "First",0;
		
	}
	cout << "Second";
	
}