杨丰磊

#include <bits/stdc++.h>
 
using namespace std;
#define N 300005
#define go(i,a,b) for(int i=(a);i<=(b);i++)
#define inf 0x3f3f3f3f
#define mod 998244353
#define ll long long
int ny(int x){return x==1?1:1ll*ny(mod%x)*(mod-mod/x)%mod; }
ll a[410][410],jx[N];
int in[N],out[N],n;
ll det(int n)//求前n行n列的行列式的值
    {
        go(i,1,n)go(j,1,n)a[i][j]=(a[i][j]%mod+mod)%mod;
        ll ret=1;
        for(int i=1;i<=n;i++)
        {
            for(int j=i+1;j<=n;j++)
                while(a[j][i])
                {
                    ll t=a[i][i]/a[j][i];
                    for(int k=i;k<=n;++k)
                        a[i][k]=((a[i][k]-a[j][k]*t)%mod+mod)%mod;
                    for(int k=i;k<=n;++k)
                        swap(a[i][k],a[j][k]);
                    ret=-ret;
                }
            if(!a[i][i])
                return 0;
            ret=ret*a[i][i]%mod;
        }
        ret=(ret%mod+mod)%mod;
        return ret;
    }
int x,cas=1;
int main()
{
    jx[0]=1;go(i,1,N-1)jx[i]=jx[i-1]*i%mod;
    while(scanf("%d",&n)!=EOF){
        ll ans=1;
        go(i,0,n)go(j,0,n)a[i][j]=out[i]=in[i]=0;
        go(i,1,n){
            go(j,1,n){
                scanf("%d",&x);
                a[i][j]-=x;
                a[j][j]+=x;
                in[i]+=x;
                out[j]+=x;
                if(x)(ans*=ny(jx[x]))%=mod;
            }
            (ans*=(jx[in[i]-1]+mod)%mod)%=mod;
        }
        int fl=1;
        go(i,1,n)if(in[i]!=out[i])fl=0;
        (ans*=det(n-1))%=mod;
       // cout<<ans<<endl;
        printf("Case #%d: %lld\n",cas++,ans*in[1]*fl%mod);
    }
    return 0;
}