題目大意：略。

解題思路：K倍動態減法，K=2的情況為斐波那契數列。

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

typedef long long ll;
int n;
ll k, fbi[50];

bool judge () {
    for (int i = 0; i < n; i++)
        if (fbi[i] == k)
            return false;
    return true;
}

int
 
 main () {
    fbi[0] = 2;
    fbi[1] = 3;
    for (n = 2; fbi[n-1] >= (1<<31); n++)
        fbi[n] = fbi[n-1] + fbi[n-2];
    while (scanf("%lld", &k) == 1 && k) {
        printf("%s\n", judge() ? "First win" : "Second win");
    }
    return 0;
}