Making Sequences is Fun373B

We'll define S(n) for positive integer n as follows: the number of the n's digits in the decimal base. For example, S(893) = 3,S(114514) = 6.

You want to make a consecutive integer sequence starting from number m (m, m + 1, ...). But you need to pay S(n)·k to add the numbern to the sequence.

You can spend a cost up to w, and you want to make the sequence as long as possible. Write a program that tells sequence's maximum length.

Input

The first line contains three integers w (1 ≤ w ≤ 1016), m (1 ≤ m ≤ 1016), k (1 ≤ k ≤ 109).

Please, do not write the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64dspecifier.

Output

The first line should contain a single integer — the answer to the problem.

Sample Input

Input

9 1 1

Output

9

Input

77 7 7

Output

7

Input

114 5 14

Output

6

Input

1 1 2

Output

0

#include <stdio.h>
#include <string.h>
long long SS(long long a)
{
long long x=0;
while(a>0)
{
a=a/10;
x++;
//printf("%I64d\n",a);
}
return x;
}

long long num(long long x)
{
long long i,y=1;
for(i=1;i<=x;i++)
y=y*10;
return y;
}
int main()
{
long long w,m,k;
int i,j;
while(scanf("%I64d %I64d %I64d",&w,&m,&k)!=EOF)
{
long long n=0,nu,ww,mm;
long long oo=SS(m);
nu=num(oo);
ww=w/k;
while(ww>0)
{
mm=ww/oo;
n=n+mm;
if(n>=nu-m)
{
mm=mm-(n-nu+m);
n=nu-m;
ww=ww-oo*mm;
oo++;
nu=num(oo);
}
else
{
ww=ww-oo*mm;
}
if(mm==0)
break;
//printf("%I64d %I64d %I64d",ww,);
}
printf("%I64d\n",n);
}
return 0;
}

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