打表可得结论：
1.只有2,3,6,7,8,11,12,15,18,19,...,108,112,128这31个数的k值是无穷大
2.当n足够大的时候，即当n>506时，设$f(x)=1^2+2^2+...+x^2=\frac{x(x+1)(2x+1)}{6}$，
找到一个t使得$f(t-1)+1\leq n\leq f(t)$，
若k(f(t)-n)是无穷大，则k(n)=t+1，否则k(n)=t
所以当n<=506时，暴力打表，否则二分查找出这个t，然后套公式即可。
#include<cstdio>
#define N 507
typedef long long ll;
ll n,l=12,r=1442250,mid,t,ans;
int i,j,v[N],sum[N],f[N]={0,1,0,0,2,2,0,0,0,3,3,0,0,3,3,0,4,4,0,0,4,4,0,0,0,4,4,0,0,4,4,0,0,0,5,5,6,6,5,5,6,5,5,0,0,5,5,0,0,6,5,5,6,6,5,5,6,6,7,7,0,6,6,7,8,6,6,0,8,7,6,6,0,8,6,6,0,6,6,7,8,6,6,7,7,7,6,6,7,7,6,6,0,8,7,7,0,9,7,7,7,7,7,7,7,7,7,9,0,8,7,7,0,8,7,7,8,8,8,7,7,8,8,7,7,8,7,7,0,8,7,7,9,8,8,7,7,9,8,7,7,8,8,8,9,8,8,8,8,8,8,8,8,8,8,8,9,10,8,8,9,9,8,8,8,8,8,8,8,8,8,9,9,10,8,8,9,10,8,8,9,9,9,8,8,9,9,8,8,10,8,8,9,10,8,8,9,9,9,8,8,9,9,8,8,9,9,9,9,10,9,9,9,10,9,9,9,9,10,9,9,9,9,9,9,10,9,9,9,9,9,9,9,9,9,9,9,10,10,9,9,10,10,9,9,9,9,9,9,9,9,9,10,10,10,9,9,11,10,9,9,10,10,10,9,9,10,10,9,9,10,9,9,11,10,9,9,11,10,10,9,9,10,10,9,9,10,10,10,11,10,10,10,11,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,11,10,10,10,11,10,10,10,10,11,10,10,10,10,10,10,11,10,10,10,10,10,10,10,10,10,10,10,11,11,10,10,11,11,10,10,10,10,10,10,10,10,10,11,11,11,10,10,11,11,10,10,11,11,11,10,10,11,11,10,10,11,10,10,11,11,10,10,11,12,11,10,10,11,11,10,10,11,11,11,11,11,11,11,11,12,11,11,11,12,11,11,11,11,11,11,11,11,11,11,11,12,11,11,11,12,11,11,11,11,11,11,11,11,11,11,11,11,11,11,11,12,11,11,11,12,11,11,11,11,12,11,11,11,11,11,11,12,11,11,11,11,11,11,11,11,11,11,11,12,12,11,11,12,12,11,11,11,11,11,11,11,11,11,12,12,12,11,11,12,12,11,11,12,12,12,11,11,12,12,11,11,12,11,11,12,12,11,11,12,12,12,11,11,12,12,11,11};
ll F(ll x){return x*(x+1)*(x*2+1)/6;}
int main(){
scanf("%lld",&n);
for(i=2;i<N;i++)if(f[i])for(j=1;j<i;j++)if(!f[j]||f[j]>f[i])v[j]=1;
for(i=2;i<N;i++)sum[i]=sum[i-1]+v[i];
if(n<N){
if(f[n])printf("%d",f[n]);else putchar('-');
return printf(" %d",sum[n]),0;
}
while(l<=r)if(F(mid=(l+r)>>1)>=n)r=(t=mid)-1;else l=mid+1;
printf("%lld ",t+(F(t)>n&&F(t)-n<=128&&!f[F(t)-n]));
for(ans=(t-12)*31+sum[N-1],i=1;i<=128;i++)if(!f[i]&&F(t)-i<=n)ans++;
return printf("%lld",ans),0;
}