World Cup

The last stage of Football World Cup is played using the play-off system.

There are n teams left in this stage, they are enumerated from 1 to n. Several rounds are held, in each round the remaining teams are sorted in the order of their ids, then the first in this order plays with the second, the third — with the fourth, the fifth — with the sixth, and so on. It is guaranteed that in each round there is even number of teams. The winner of each game advances to the next round, the loser is eliminated from the tournament, there are no draws. In the last round there is the only game with two remaining teams: the round is called the Final, the winner is called the champion, and the tournament is over.

Arkady wants his two favorite teams to play in the Final. Unfortunately, the team ids are already determined, and it may happen that it is impossible for teams to meet in the Final, because they are to meet in some earlier stage, if they are strong enough. Determine, in which round the teams with ids a and b can meet.

Input

The only line contains three integers n, a and b (2 ≤ n ≤ 256, 1 ≤ a, b ≤ n) — the total number of teams, and the ids of the teams that Arkady is interested in.

It is guaranteed that n is such that in each round an even number of team advance, and that a and b are not equal.

Output

In the only line print "Final!" (without quotes), if teams a and b can meet in the Final.

Otherwise, print a single integer — the number of the round in which teams a and b can meet. The round are enumerated from 1.

Examples

Input

4 1 2

Output

1

Input

8 2 6

Output

Final!

Input

8 7 5

Output

2

Note

In the first example teams 1 and 2 meet in the first round.

In the second example teams 2 and 6 can only meet in the third round, which is the Final, if they win all their opponents in earlier rounds.

In the third example the teams with ids 7 and 5 can meet in the second round, if they win their opponents in the first round.

这个题好玩啊，就好比梅西和c罗会在那一场遇到一样（当然你也知道踢球是两个队伍取一个），然后你一看代码，就明白其实他是一个判断多少轮之后两个队伍的值相等，然后再根据多少轮看看是否超出比赛队伍总场数的二倍。

#include
#include
using namespace std;
int main()
{
    int n,a,b;
    while(~scanf("%d%d%d",&n,&a,&b))
    {
        int winner,loster,lun=1,sum=0;
        winner=min(a,b);
        loster=max(a,b);
        while(winner!=loster)
        {
            winner=(winner+1)/2;
            loster=(loster+1)/2;
            lun*=2;
            sum++;
        }
        if(lun>=n)
            printf("Final!\n");
        else
            printf("%d\n",sum);
    }
    return 0;
}