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>
#define fr(i, n) for (int i = 0; i < n; ++i)
#define frab(i, a, b) for (int i = a; i < b; ++i)
#define mp make_pair
#define pb push_back

using namespace std;

typedef long long ll;
typedef double ld;

const int N = 1e5 + 10;
const int INF = 1e9 + 10;

vector <int> g[N];

int sz[N];
int n;
bool used[N];
bool fnd = false;

void dfs(int v, int par) {
    //used[v] = true;
    for (auto t: g[v])
        if (t != par)
            dfs(t, v);
    int sum = 0;
    int odd = 0;
    for (auto t: g[v])
        if (t != par) {
            sum += sz[t];
            if (sz[t] % 2)
                odd++;
        }
    sz[v] = sum + 1;
    if ((n - sum - 1) % 2)
        odd++;
    if (odd >= 2) fnd = true;
}

int main() {
    //freopen("a.in", "r", stdin);
    cin >> n;
    fr(i, n - 1) {
        int v, u;
        cin >> v >> u;
        g[--v].pb(--u);
        g[u].pb(v);
    }

    dfs(0, -1);
    if (fnd)
        cout << "First";
    else
        cout << "Second";
}