codeforce A. Accounting

A. Accounting

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

A long time ago in some far country lived king Copa. After the recent king's reform, he got so large powers that started to keep the books by himself.

The total income A of his kingdom during 0-th year is known, as well as the total income B during n-th year (these numbers can be negative — it means that there was a loss in the correspondent year).

King wants to show financial stability. To do this, he needs to find common coefficient X — the coefficient of income growth during one year. This coefficient should satisfy the equation:

A·Xn = B.

Surely, the king is not going to do this job by himself, and demands you to find such number X.

It is necessary to point out that the fractional numbers are not used in kingdom's economy. That's why all input numbers as well as coefficient X must be integers. The number X may be zero or negative.

Input

The input contains three integers A, B, n (|A|, |B| ≤ 1000, 1 ≤ n ≤ 10).

Output

Output the required integer coefficient X, or «No solution», if such a coefficient does not exist or it is fractional. If there are several possible solutions, output any of them.

Examples

input

Copy

2 18 2

output

3

input

Copy

-1 8 3

output

-2

input

Copy

0 0 10

output

5

input

Copy

1 16 5

output

No solution



水题wa了好多遍，思维还是不行 （对于0的讨论，上代码了

#include <bits/stdc++.h>
using namespace std;

int main() {
int a,b,n;
cin>>a>>b>>n;
if(a==0&&b==0)
{
printf("0 ");
return 0;
}
else if(a==0&&b!=0)
{
printf("No solution ");
return 0;
}
else if(a!=0&&b==0)
{
printf("0 ");
return 0;
}
int temp = b/a;
bool flag=true;
if(temp<0)
{
flag=false;
temp=-temp;
}
for(int i = 1;i<=temp;i++)
{
long long int tem= 1;
for(int j = 1;j<=n;j++ )
{
tem*=i;
}
// printf("%d ",tem);
if(tem==temp)
{
if(flag)printf("%d ",i);
else {
printf("%d ",-i);
}
return 0;
}
}
printf("No solution ");
return 0;
}