CodeForces 373B——模拟——Making Sequences is Fun

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<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
long long  w, m , k;
long long temp1(long long x)
{
long long y = 0;
while(x > 0){
x /= 10;
y++;
}
return y;
}

long long temp2(long long x)
{
long long num = 1;
for(int i = 1; i <= x; i++)
num *= 10;
return num;
}

int main()
{
long long sum;
while(~scanf("%lld%lld%lld", &w, &m, &k)){
long long a = temp1(m);//位数
long long t = w / k;//需要减掉的数
long long t1 = t / a;//所需要的数的个数
long long t2 = temp2(a);//10.....
long long ans = 0;
while(t > 0){
if(t1 > t2 - m){
t -= (t2 - m) * a;
ans += t2 - m;
// printf("%d\n",ans);
a++;
t1= t / a;
t2 = temp2(a);
m = temp2(a-1);
}
else {
t -= t1 * a;
ans += t1;
break;
}
}
printf("%lld\n", ans);
}
return 0;
}