#include<cstdio>
#include<iostream>
#include<vector>
#include<algorithm>
#include<string>
#include<cstring>
using namespace std;

typedef long long LL;
typedef vector<int> VI;

#define REP(i,n) for(int i=0, i##_len=(n); i<i##_len; ++i)
#define EACH(i,c) for(__typeof((c).begin()) i=(c).begin(),i##_end=(c).end();i!=i##_end;++i)
#define eprintf(s...) fprintf(stderr, s)

template<class T> inline void amin(T &a, const T &b) { if (b<a) a=b; }
template<class T> inline void amax(T &a, const T &b) { if (a<b) a=b; }


int N, M, A[1011][1011], S[1011];
double dp[1011];

int main() {
    for (;scanf("%d%d", &N, &M) != EOF;) {
    REP (i, N) REP (j, M) scanf("%d", &A[i][j]);

#ifdef IGNORE
    N = 1000; M = 1000;
    REP (i, N) REP (j, M) A[i][j] = rand() % 100 + 1;
#endif

    REP (i, N) {
        S[i] = 0;
        REP (j, M) S[i] += A[i][j];
    }

    double ans = 0;
    REP (j, M) {
        memset(dp, 0, sizeof dp);
        dp[0] = 1;
        REP (i, N) {
        double p = (double)A[i][j] / S[i];
        for (int k=min(20, i+1); k>0; k--) {
            dp[k] = dp[k] * (1-p) + dp[k-1] * p;
        }
        dp[0] *= (1-p);
        }

        REP (i, N+1) ans += i*i*dp[i];
    }

    printf("%.2f\n", ans);
    }
    return 0;
}