#include <bits/stdc++.h> using namespace std; // #define int long long #define rep(i, n) for (long long i = (long long)(0); i < (long long)(n); ++i) #define reps(i, n) for (long long i = (long long)(1); i <= (long long)(n); ++i) #define rrep(i, n) for (long long i = ((long long)(n)-1); i >= 0; i--) #define rreps(i, n) for (long long i = ((long long)(n)); i > 0; i--) #define irep(i, m, n) for (long long i = (long long)(m); i < (long long)(n); ++i) #define ireps(i, m, n) for (long long i = (long long)(m); i <= (long long)(n); ++i) #define SORT(v, n) sort(v, v + n); #define REVERSE(v, n) reverse(v, v+n); #define vsort(v) sort(v.begin(), v.end()); #define all(v) v.begin(), v.end() #define mp(n, m) make_pair(n, m); #define cout(d) cout<<d<<endl; #define coutd(d) cout<<std::setprecision(10)<<d<<endl; #define cinline(n) getline(cin,n); #define replace_all(s, b, a) replace(s.begin(),s.end(), b, a); #define PI (acos(-1)) #define FILL(v, n, x) fill(v, v + n, x); #define sz(x) long long(x.size()) using ll = long long; using vi = vector<int>; using vvi = vector<vi>; using vll = vector<ll>; using vvll = vector<vll>; using pii = pair<int, int>; using pll = pair<ll, ll>; using vs = vector<string>; using vpll = vector<pair<ll, ll>>; using vtp = vector<tuple<ll,ll,ll>>; using vb = vector<bool>; template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; } template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; } const ll INF = 1e9+10; const ll MOD = 1e9+7; const ll LINF = 1e18; ll dp[5005][15005]; // i番目まででj問解いた時の順位の和の最小値 signed main() { cin.tie( 0 ); ios::sync_with_stdio( false ); ll n,p; cin>>n>>p; rep(i,n+1) rep(j,p+1) dp[i][j]=LINF; dp[0][0]=0; rep(i,n){ ll a,b,c; cin>>a>>b>>c; for(ll j=0; j<=i*3; j++){ if(dp[i][j]==LINF) continue; chmin(dp[i+1][j], dp[i][j]+a); chmin(dp[i+1][j+1], dp[i][j]+b); chmin(dp[i+1][j+2], dp[i][j]+c); chmin(dp[i+1][j+3], dp[i][j]+1); } } // rep(i,p+1) cout<<dp[n][i]<<' '; // cout<<endl; double ans=1.0*dp[n][p]/n; cout<<std::setprecision(10)<<ans<<endl; }