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/*
动态转移方程f[i][j]=(f[i+1][j-1]+g[i-1][j-1])*l[i][z]*r[j][z];
f[i][j]是区间[i+1,j-1]匹配的概率与i,j处匹配的概率；
g[i][j]是区间[i,j]以()(())...()形式匹配的概率；
答案f[1][n]+g[1][n];
*/
#include<cstdio>
int n,k;
double f[201][201],g[201][201],l[201][51],r[201][51];
int main(){
    freopen("brackets.in","r",stdin);
    freopen("brackets.out","w",stdout);
    scanf("%d%d",&n,&k);
    if(n==0){printf("1.00000\n");fclose(stdin);fclose(stdout);return 0;}
    for(int i=1;i<=n;i++)
    for(int j=1;j<=k;j++) scanf("%lf%lf",&l[i][j],&r[i][j]);
    for(int i=0;i<=n;i++) f[i+1][i]=1;
    for(int len=2;len<=n;len+=2)
    for(int i=1;i<=n-len+1;i++){
        for(int as=1;as<=k;as++)
            f[i][i+len-1]+=(f[i+1][i+len-2]+g[i+1][i+len-2])*l[i][as]*r[i+len-1][as];
        for(int mid=i+1;mid<=i+len-2;mid+=2)
            g[i][i+len-1]+=f[i][mid]*(f[mid+1][i+len-1]+g[mid+1][i+len-1]);
    }
    printf("%.5lf\n",f[1][n]+g[1][n]);
    fclose(stdin);
    fclose(stdout);
    return 0;
}