[title link]
　　http://www.lydsy.com/JudgeOnline/problem.php?id=1070
[Abstract]
The way to build a map is still quite magical.
We don't consider the waiting cost before each car, but consider the impact of each car on the next car, so we put each car in the front. The cost is the number of cars behind it × its time. It is the number of cars behind it × its time.
Therefore, each technician is divided into nn points, and the flow at each point is limited to 11, which means the vehicle repaired in the second to last place.
Each car also builds a point to connect with each time of each counting person. The cost is i * costi * cost

/* --------------
    user Vanisher
    problem
----------------*/
# include <bits/stdc++.h>
# define    ll      long long
# define    inf     0x3f3f3f3f
# define    M       100010
# define    N       1010
# define    MM      20
# define    NN      100
using namespace std;
int read(){
    int tmp=0, fh=1; char ch=getchar();
    while (ch<'0'||ch>'9'){if (ch=='-') fh=-1; ch=getchar();}
    while (ch>='0'&&ch<='9'){tmp=tmp*10+ch-'0'; ch=getchar();}
    return tmp*fh;
}
struct node{
    int data,next,l,vote,re;
}e[M];
int place,head[N],dis[N],use[N],can[N],ned[N],frm[N],n,m,S,T,q[N],ti;
int mp[MM][NN],pin[MM][NN],p[NN];
ll ans;
void build(int u, int v, int l, int w){
    e[++place].data=v; e[place].next=head[u]; head[u]=place; e[place].vote=w; e[place].l=l; e[place].re=place+1;
    e[++place].data=u; e[place].next=head[v]; head[v]=place; e[place].vote=-w; e[place].l=0; e[place].re=place-1;
}
void spfa(){
    memset(dis,inf,sizeof(dis));
    memset(use,0,sizeof(use));
    dis[S]=0; use[S]=1; can[S]=inf;
    int pl=1, pr=1; q[1]=S;
    while (pl<=pr){
        int x=q[(pl++)%N];
        for (int ed=head[x]; ed!=0; ed=e[ed].next)
            if (dis[e[ed].data]>dis[x]+e[ed].vote&&e[ed].l!=0){
                dis[e[ed].data]=dis[x]+e[ed].vote;
                can[e[ed].data]=min(e[ed].l,can[x]);
                frm[e[ed].data]=ed;
                if (use[e[ed].data]==0){
                    use[e[ed].data]=1;
                    q[(++pr)%N]=e[ed].data;
                }
            }
        use[x]=0;
    }
}
void change(){
    ans=ans+(long long)dis[T]*(long long)can[T];
    int now=T;
    while (now!=S){
        e[frm[now]].l-=can[T];
        e[e[frm[now]].re].l+=can[T];
        now=e[e[frm[now]].re].data;
    }
}
void flow(){
    for (spfa(); dis[T]!=inf; spfa())
        change();
}
int main(){
    m=read(), n=read();
    S=0; T=1; ti=1;
    for (int i=1; i<=n; i++)
        for (int j=1; j<=m; j++){
            mp[j][i]=read();
            pin[j][i]=++ti;
            build(pin[j][i],T,1,0);
        }
    for (int i=1; i<=n; i++){
        p[i]=++ti;
        build(S,p[i],1,0);
        for (int j=1; j<=m; j++)
            for (int k=1; k<=n; k++)
                build(p[i],pin[j][k],1,mp[j][i]*k);
    }
    flow();
    printf("%.2lf\n",ans*1.0/n);
    return 0;
}