A. Tile Painting

A. Tile Painting

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Ujan has been lazy lately, but now has decided to bring his yard to good shape. First, he decided to paint the path from his house to the gate.

The path consists of nn consecutive tiles, numbered from 11 to nn. Ujan will paint each tile in some color. He will consider the path aesthetic if for any two different tiles with numbers ii and jj, such that |j−i||j−i| is a divisor of nn greater than 11, they have the same color. Formally, the colors of two tiles with numbers ii and jj should be the same if |i−j|>1|i−j|>1 and nmod|i−j|=0nmod|i−j|=0 (where xmodyxmody is the remainder when dividing xx by yy).

Ujan wants to brighten up space. What is the maximum number of different colors that Ujan can use, so that the path is aesthetic?

Input

The first line of input contains a single integer nn (1≤n≤10121≤n≤1012), the length of the path.

Output

Output a single integer, the maximum possible number of colors that the path can be painted in.

Examples

input

Copy

4

output

Copy

2

input

Copy

5

output

Copy

5

Note

In the first sample, two colors is the maximum number. Tiles 11 and 33 should have the same color since 4mod|3−1|=04mod|3−1|=0. Also, tiles 22 and 44 should have the same color since 4mod|4−2|=04mod|4−2|=0.

In the second sample, all five colors can be used.

 还是说找规律，写2-10的样例，发现质数一定是本身数量，合数大部分是1,4是2,9是3,也就可以猜测出平方数就是其开方的结果，交上去发现WA了，又验证16，25.发现16并非如此，只有25才是这样。

# include 
#include
# include
# include
# include
using namespace std;
typedef long long int ll;
 
int main()
{
 
    ll n;
 
    cin>>n;
 
    ll x=n;
    ll cnt=0,flag=0,ans;
    for(ll i=2;i*i<=x;i++)
    {
        if(x%i==0)
        {
           while(x%i==0)
           {
               x/=i;
           }
 
           ans=i;
 
           cnt++;
        }
    }
 
    if(x>1)
        {
            cnt++;
            flag=1;
        }
        
        
 
    if(cnt==1&&flag)
    {
        cout<
    }
    else
    {
        if(cnt==1)
        {
            cout<
        }
        else
        {
            cout<<1;
        }
    }
 
 
    return  0;
}

那就猜测质因数是1个的情况，交上去发现A了。

可以说C题的数论主要还是GCD与质因数分解，往这方面多想想就行