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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>

typedef struct {
	int parent;
	int exist;
}node;

int main(){
	int N, a, b, i, tmp;
	scanf("%d", &N);
	node *tree = (node *)malloc(sizeof(node) * (N + 1));
	tree[0].parent = -1;
	tree[0].exist = -1;
	for(i = 1; i <= N; i++){
		tree[i].exist = 1;
	}
	tree[1].parent = 0;
	for(i = 1; i < N; i++){
		scanf("%d%d", &a, &b);
		if(a > b){
			tmp = a;
			a = b;
			b = tmp;
		}
		tree[b].parent = a;
	}
	if(N % 2 == 1){
		printf("First\n");
		return 0;
	}
	int orphan = 0;
	for(i = N; i > 0; i--){
		if(tree[i].exist == 1){
			if(tree[tree[i].parent].exist == 1){
				tree[i].exist = -1;
				tree[tree[i].parent].exist = -1;
			}
			else{
				orphan++;
			}
		}
	}
	if(orphan > 0){
		printf("First\n");
	}
	else{
		printf("dummy\n");
		printf("Second\n");
	}
	return 0;
}