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 <iostream>
#include <vector>

struct Node {
	std::vector<int> edgeTo;
	int deadEndNum = 0;
};

int main() {
	int n;
	std::cin >> n;

	std::vector<Node> nodes(n);

	for (int i = 0; i < n - 1; ++i) {
		int a, b;
		std::cin >> a >> b;
		--a, --b;

		nodes[a].edgeTo.push_back(b);
		nodes[b].edgeTo.push_back(a);
	}

	for (auto&& node : nodes)
		if (node.edgeTo.size() == 1)
			++nodes[node.edgeTo.front()].deadEndNum;

	for (int i = 0; i < n; ++i) {
		if (nodes[i].deadEndNum >= 2) {
			std::cout << "First" << std::endl;
			return 0;
		}

		if (nodes[i].edgeTo.size() == 1) {
			int prevD = i;
			int d = nodes[i].edgeTo.front();

			for (int j = 0; j < 2; ++j) {
				const auto& to = nodes[d].edgeTo;

				if (to.size() == 2) {
					int tmp = prevD;
					prevD = d;
					d = (to[0] == tmp ? to[1] : to[0]);
				}
				else {
					d = -1;
					break;
				}
			}
			if (d != -1 && nodes[d].edgeTo.size() >= 2) {
				std::cout << "First" << std::endl;
				return 0;
			}
		}
	}

	std::cout << "Second" << std::endl;
}