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 Kotlin (1.0.0) (judge verdict: Wrong Answer)

import java.util.*

fun main(args: Array<String>){
	val cin = Scanner(System.`in`)
    val n = cin.nextInt()

    var arr = Array<MutableList<Int>>(n, {mutableListOf()})

    for (i in 1..n-1) {
        val a = cin.nextInt() - 1
        val b = cin.nextInt() - 1
        arr[a].add(b)
        arr[b].add(a)
    }

    var start: Int = -1

    for (i in 0..n-2) {
        if (arr[i].size == 1) {
            start = i
            break
        }
    }

    val odd = Array<Int>(n, {-1})
    odd[start] = 1
    val queue: MutableList<Int> = mutableListOf(start)

    var f = false
    
    while(queue.size != 0) {
        val now = queue.removeAt(0)
        for (i in arr[now]) {
            if (odd[i] == -1) {
                odd[i] = (odd[now]+1) % 2
                queue.add(i)
            }
        }
        if (now != start && arr[now].size == 1 && odd[now] == 1) {
            f = true
            break
        }
    }
    if (f) {
        println("First")
    } else {
        println("Second")
    }
}