#382 div.2

• Tennis Championship
  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.

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

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

Example
Input
2
Output
1
Input
3
Output
2
Input
4
Output
2
Input
10
Output
4
Note
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.

啊咧= =要注意类似这种题最好不过定性分析
之前的呢有一道母牛的题
牛的数量=现在牛的数量+能生牛的数量
同理
这道题：
比赛的局数=冠军和他将要比赛的人

//AC代码：
#include <iostream>
using namespace std;
int main()
{
    long long  n,f[1000];
    cin>>n;
    f[1]=1;
    f[2]=2;
    int b=2;
    while(n>=f[b]){
        ++b;
        f[b]=f[b-1]+f[b-2];
    }

    cout<<b-2<<endl;
}