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;

int N, col1, col2;
bool meow[100005];
vector<int> V[100005];

void dfs(int x, int p, int col) {
	if (col == 0) col1++;
	else col2++;
	for (int i = 0; i < V[x].size(); i++) if (p != V[x][i]) dfs(V[x][i], x, !col);
}

int main() {
	scanf("%d", &N);
	for (int i = 0; i < N - 1; i++) {
		int a, b;
		scanf("%d%d", &a, &b);
		a--;
		b--;
		V[a].push_back(b);
		V[b].push_back(a);
	}
	for (int i = 0; i < N; i++) {
		int leafs = 0;
		for (int j = 0; j < V[i].size(); j++) if (V[V[i][j]].size() == 1) leafs++;
		if (leafs > 1) {
			printf("First\n");
			return 0;
		} else if (leafs == 1) meow[i] = 1;
	}
	for (int i = 0; i < N; i++) {
		if (V[i].size() == 2 && meow[V[i][0]] == 1 && meow[V[i][1]] == 1) {
			printf("First\n");
			return 0;
		}
	}
	dfs(0, -1, 0);
	if (col1 == col2) printf("Second\n");
	else printf("First\n");
}