Famous Brazil city Rio de Janeiro holds a tennis tournament and Ostap Bender doesn't want to miss this event. There will be n players
participating, and the tournament will follow knockout rules from the very first game. That means, that if someone loses a game he leaves the tournament immediately.

Organizers are still arranging tournament grid (i.e. the order games will happen and who is going to play with whom) but they have already fixed one rule: two players can play against each other only if the number of games one of them has already played differs
by no more than one from the number of games the other one has already played. Of course, both players had to win all their games in order to continue participating in the tournament.

Tournament hasn't started yet so the audience is a bit bored. Ostap decided to find out what is the maximum number of games the winner of the tournament can take part in (assuming the rule above is used). However, it is unlikely he can deal with this problem
without your help.

The only line of the input contains a single integer n (2 ≤ n ≤ 1018) —
the number of players to participate in the tournament.

Print the maximum number of games in which the winner of the tournament can take part.

In all samples we consider that player number 1 is the winner.

In the first sample, there would be only one game so the answer is 1.

In the second sample, player 1 can consequently beat players 2 and 3.

In the third sample, player 1 can't play with each other player as after he plays with players 2 and 3 he
can't play against player 4, as he has 0 games
played, while player 1 already played 2.
Thus, the answer is 2 and to achieve we make pairs (1, 2) and (3, 4) and
then clash the winners.

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题目的意思是n个人进行比赛，输了直接淘汰，而每个人只能和赢的场数和自己不超过1的人打，问最后比赛最多次的人比赛了几场？

我们可以清楚 如果一个人要赢k场，那么他必须先赢k-1场，同时在打败一个赢了k-2场的人，所以我们可以清楚地发现这是一个斐波那契数列

所以我们先打表，然后找到小于等于n的是第几位就好了

#include <iostream>

#include <cstdio>

#include <cstring>

#include <algorithm>

#include <cmath>

using namespace std;

int main()

{

long long  n;

long long int a[91];

a[0]=0;

 a[1]=2;

 a[2]=3;

 for(int i=3;i<=90;i++)

    a[i]=a[i-1]+a[i-2];

while(~scanf("%I64d",&n))

{

    int cnt=lower_bound(a,a+90,n)-a;

    cnt--;

    for(int i=1;i<=90;i++)

        if(n==a[i])

        cnt++;

    cout<<cnt<<endl;

}

    return 0;

}