#include <bits/stdc++.h>
using namespace std;
/*{{{*/  //template
#define rep(i,n) for(int i=0;i<n;i++)
constexpr int INF = numeric_limits<int>::max()/2;
constexpr long long LINF = numeric_limits<long long>::max()/3;
#define mp make_pair
#define pb push_back
#define eb emplace_back
#define fi first
#define se second
#define all(v) (v).begin(),(v).end()
#define sz(x) (int)(x).size()
#define debug(x) cerr<<#x<<":"<<x<<endl
#define debug2(x,y) cerr<<#x<<","<<#y":"<<x<<","<<y<<endl
//struct fin{ fin(){ cin.tie(0); ios::sync_with_stdio(false); } } fin_;
struct Double{ double d; explicit Double(double x) : d(x){} };
ostream& operator<<(ostream& os,const Double x){ os << fixed << setprecision(20) << x.d; return os; }
template<typename T> ostream& operator<<(ostream& os,const vector<T>& vec){ os << "["; for(const auto& v : vec){ os << v << ","; } os << "]"; return os; }
template<typename T,typename U> ostream& operator<<(ostream& os,const pair<T,U>& p){ os << "(" << p.first << ","<< p.second <<")"; return os; }
template<typename T> ostream& operator<<(ostream& os,const set<T>& st){ os<<"{"; for(T v:st) os<<v<<","; os <<"}"; return os; }
template<typename T,typename U> inline void chmax(T &x,U y){ if(y>x) x = y; }
template<typename T,typename U> inline void chmin(T &x,U y){ if(y<x) x = y; }
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
typedef vector<int> vi;
typedef vector<vi> vvi;
ll gcd(ll a,ll b){ if(b==0) return a; else return gcd(b,a%b); }
//constexpr double eps = 1e-14; 
constexpr double eps = 1e-10; 
constexpr ll mod = 1e9+7;
const int dx[]={1,0,-1,0} ,dy[] = {0,1,0,-1};
/*}}}*/


constexpr ll max_n = 5000;
int N,P;
int r[max_n][4];
int dp[max_n+1][max_n*3+1];

int main(){
    cin >> N >> P;
    rep(i,N){
        rep(j,3) cin >> r[i][j];
        r[i][3] = 1;
    }
    rep(i,N+1) rep(j,P+1){
        //cout << i <<  " " << j << endl;
        dp[i][j] = INF;
    }
    dp[0][0] = 0;

    for(int i=0;i<N;i++){   // solve r[i] -> update dp[i+1]
        for(int j=0;j<=P;j++){
            for(int k=0;k<4;k++){   // k問解く
                int nj = j + k;
                if(nj > P) break;

                dp[i+1][nj] = min(dp[i+1][nj] , dp[i][j] + r[i][k]);
            }
        }
    }

    cout << Double(dp[N][P] / (double)N) << endl;
}