猴子之间的打架是棵无根树，有\(n^{n-2}\)种可能；同时n-1个过程的排列是\((n-1)!\)
//820kb 104ms
#include <cstdio>
const int mod=9999991;
int FP(long long x,int k)
{
long long t=1;
for(; k; k>>=1,x=x*x%mod)
if(k&1) t=t*x%mod;
return t;
}
int main()
{
int n; scanf("%d",&n);
long long res=1;
for(int i=2; i<n; ++i) res*=i, res>=mod?res%=mod:0;
printf("%d",(int)(FP(n,n-2)*res%mod));
return 0;
}