CodeForces 569A Music

D - Music
Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d
& %I64u
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569A

Appoint description: 
System Crawler  (2016-06-20)

Description

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 is T 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. For q seconds of
real time the Internet allows you to download q - 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 Input

Input

5 2 2

Output

2

Input

5 4 7

Output

1

Input

6 2 3

Output

1

Hint

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.
题意：Little Lesha 想听一首歌， 这首歌长t秒，Little Lesha 先下载了s秒，边听边下载，每q秒可以下载q-1秒的歌，每当播放到未下载的时候，返回开始从头开始播放，问Little Lesha 下载一首歌需要从头开始多少次。

思路：假定放到未下载点时候用时x秒，那么（q-1)/q就是每秒下载的速度，（q-1)/q*x+s=x。那么x=q*s

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int main()
{
int t,s,q,ans;
while(scanf("%d%d%d",&t,&s,&q)!=EOF)
{
ans=0;
while(s<t)
{
s=s*q;
ans++;
}
printf("%d\n",ans);
}
return 0;
}