題干：

Valeric and Valerko missed the last Euro football game, so they decided to watch the game's key moments on the Net. They want to start watching as soon as possible but the connection speed is too low. If they turn on the video right now, it will "hang up" as the size of data to watch per second will be more than the size of downloaded data per second.

The guys want to watch the whole video without any pauses, so they have to wait some?integer?number of seconds for a part of the video to download. After this number of seconds passes, they can start watching. Waiting for the whole video to download isn't necessary as the video can download after the guys started to watch.

Let's suppose that video's length is?c?seconds and Valeric and Valerko wait?tseconds before the watching. Then for any moment of time?t0,?t?≤?t0?≤?c?+?t, the following condition must fulfill: the size of data received in?t0?seconds is not less than the size of data needed to watch?t0?-?t?seconds of the video.

Of course, the guys want to wait as little as possible, so your task is to find the minimum integer number of seconds to wait before turning the video on. The guys must watch the video without pauses.

Input

The first line contains three space-separated integers?a,?b?and?c?(1?≤?a,?b,?c?≤?1000,?a?>?b). The first number (a) denotes the size of data needed to watch one second of the video. The second number (b) denotes the size of data Valeric and Valerko can download from the Net per second. The third number (c) denotes the video's length in seconds.

Output

Print a single number — the minimum integer number of seconds that Valeric and Valerko must wait to watch football without pauses.

Examples

Input

4 1 1

Output

3

Input

10 3 2

Output

5

Input

13 12 1

Output

1

Note

In the first sample video's length is 1 second and it is necessary 4 units of data for watching 1 second of video, so guys should download 4?·?1 = 4 units of data to watch the whole video. The most optimal way is to wait 3 seconds till 3 units of data will be downloaded and then start watching. While guys will be watching video 1 second, one unit of data will be downloaded and Valerik and Valerko will have 4 units of data by the end of watching. Also every moment till the end of video guys will have more data then necessary for watching.

In the second sample guys need 2?·?10 = 20 units of data, so they have to wait 5 seconds and after that they will have 20 units before the second second ends. However, if guys wait 4 seconds, they will be able to watch first second of video without pauses, but they will download 18 units of data by the end of second second and it is less then necessary.

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題目大意：

? ?輸入a,b,c，第一個(gè)數(shù)字(a)表示觀看一秒鐘視頻所需的數(shù)據(jù)大小。第二個(gè)數(shù)字(b)表示每秒可以從網(wǎng)絡(luò)下載的數(shù)據(jù)大小。第三個(gè)數(shù)字(c)表示視頻的長(zhǎng)度(以秒為單位)。問(wèn)我如果想中間不停頓的播放，則至少需要等他下載多少秒再開(kāi)始播放？

解題報(bào)告：

? ? 菜啊，，想錯(cuò)式子了啊剛開(kāi)始。一直想推一個(gè)追及的式子，結(jié)果用時(shí)間的等式或者用下載量的等式，最后都推出來(lái)類(lèi)似1=1的東西，，，崩潰，，后來(lái)發(fā)現(xiàn)不能這么看，因?yàn)樘崆皫酌胂螺d的時(shí)間你是不知道的啊。其實(shí)設(shè)提前下載t秒，一共視頻是c秒，這個(gè)告訴你了，所以t*b+c*b=a*c就可以了。。。。并且需要向上取整。。。水題

AC代碼：

#include<bits/stdc++.h> using namespace std;int main() {int a,b,c;double x;cin>>a>>b>>c;x=a*c-b*c;printf("%d\n",(int)ceil(x/b));return 0 ; }

?

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