ACM 歌曲边下边听问题

Little Lesha loves listening to music via his smartphone. But the smartphone doesn't have much memory, so Lesha listens to his favorite songs in a well-known social network InTalk.

Unfortunately, internet is not that fast in the city of Ekaterinozavodsk and the song takes a lot of time to download. But Lesha is quite impatient. The song's duration isT seconds. Lesha downloads the first S seconds of the song and plays it. When the playback reaches the point that has not yet been downloaded, Lesha immediately plays the song from the start (the loaded part of the song stays in his phone, and the download is continued from the same place), and it happens until the song is downloaded completely and Lesha listens to it to the end. Forq seconds of real time the Internet allows you to downloadq - 1 seconds of the track.

Tell Lesha, for how many times he will start the song, including the very first start.

Input

The single line contains three integers T, S, q (2 ≤ q ≤ 10⁴,1 ≤ S < T ≤ 10⁵).

Output

Print a single integer — the number of times the song will be restarted.

Sample test(s)

Input

5 2 2

Output

2

Input

5 4 7

Output

1

Input

6 2 3

Output

1

Note

In the first test, the song is played twice faster than it is downloaded, which means that during four first seconds Lesha reaches the moment that has not been downloaded, and starts the song again. After another two seconds, the song is downloaded completely, and thus, Lesha starts the song twice.

In the second test, the song is almost downloaded, and Lesha will start it only once.

In the third sample test the download finishes and Lesha finishes listening at the same moment. Note that song isn't restarted in this case.

 解题思路：

题目大意是一个人下载歌，每q个时间单位能下载q－1个时间单位的歌，歌的长度为T，下到S的时候开始播放，如果歌还没下完且放到了还未下载的地方，则重头开始放，问一共要放多少次。一开始从s开始，设下了x秒后听和下的进度相同，假设放到未下载点的时候用时x秒，每秒的下载速度（q-1）/q,x=s+x*(q-1)/q,则x=s*q。解得x= q * s，然后从头开始，设t'秒后进度相同，则(q - 1) / q * t' +x = t'，解得t' = x * q，可见直接拿第一次进度相同的时间乘q就是接下来每次进度相同的时间.

程序代码：

#include <iostream>
using namespace std;
int main()
{
    int t,s,q,n=1;
    cin>>t>>s>>q;
    int x=q*s;
    while(x<t)
    {
        x=q*x;
        n++;
    }
    cout<<n<<endl;
    return 0;
}