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 <bits/stdc++.h>

using namespace std;

enum
{
  BLACK, WHITE
};

int N;
vector< int > g[100000];
int sub[100000];
int pa[100000];

int dfs(int idx, int par = -1)
{
  sub[idx]++;
  int visited = 0;
  for(auto &to : g[idx]) {
    if(par == to) continue;
    visited++;
  }
  if(visited == 0) return (BLACK);
  int bb = 0;
  for(auto &to : g[idx]) {
    if(par == to) continue;
    int get = dfs(to, idx);
    sub[idx] += sub[to];
    if(get == BLACK) ++bb;
  }
  if(bb >= 2) {
    cout << "First" << endl;
    exit(0);
  }

  if(sub[idx] % 2 == 0) return (WHITE);
  else return (BLACK);
}

int main()
{


  cin >> N;
  for(int i = 1; i < N; i++) {
    int a, b;
    cin >> a >> b;
    --a, --b;
    g[a].push_back(b);
    g[b].push_back(a);
  }

  if(N == 2) {
    cout << "Second" << endl;
    return (0);
  }

  if(dfs(0) == BLACK) {
    cout << "First" << endl;
  } else {
    cout << "Second" << endl;
  }

}