Problem: N people run a marathon. There are M resting places on the way. For each resting place, the i-th runner takes a break with probability P_i percent. When the i-th runner takes a break, he gets rest for T_i time.

The i-th runner runs at constant speed V_i, and the distance of the marathon is L.

You are requested to compute the probability for each runner to win the first place. If a runner arrives at the goal with another person at the same time, they are not considered to win the first place.



Input

A dataset is given in the following format:

N M L
P_1 T_1 V_1
P_2 T_2 V_2
...
P_N T_N V_N


The first line of a dataset contains three integers N (1 \leq N \leq 100), M (0 \leq M \leq 50) and L (1 \leq L \leq 100,000). N is the number of runners. M is the number of resting places. L is the distance of the marathon.

Each of the following N lines contains three integers P_i (0 \leq P_i \leq 100), T_i (0 \leq T_i \leq 100) and V_i (0 \leq V_i \leq 100) describing the i-th runner. P_i is the probability to take a break. T_i is the time of resting. V_i is the speed.

Output

For each runner, you should answer the probability of winning. The i-th line in the output should be the probability that the i-th runner wins the marathon. Each number in the output should not contain an error greater than 10^{-5}.

Examples

Input

2 2 50
30 50 1
30 50 2


Output

0.28770000
0.71230000


Input

2 1 100
100 100 10
0 100 1


Output

0.00000000
1.00000000


Input

3 1 100
50 1 1
50 1 1
50 1 1


Output

0.12500000
0.12500000
0.12500000


Input

2 2 50
30 0 1
30 50 2


Output

0.51000000
0.49000000

Solution (in CPP):

Solution:
#define _CRT_SECURE_NO_WARNINGS
//#define _GLIBCXX_DEBUG
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef pair<int,int> pii;
#define all(c) (c).begin(), (c).end()
#define loop(i,a,b) for(ll i=a; i<ll(b); i++)
#define rep(i,b) loop(i,0,b)
#define pb push_back
#define eb emplace_back
#define mp make_pair
#define mt make_tuple
template<class T> istream & operator>>(istream & is, vector<T> &);
template<class T> ostream & operator<<(ostream & os, vector<T> const &);
template<int n, class...T> typename enable_if<(n>=sizeof...(T))>::type _ot(ostream &, tuple<T...> const &){}
template<int n, class...T> typename enable_if<(n< sizeof...(T))>::type _ot(ostream & os, tuple<T...> const & t){ os << (n==0?"":" ") << get<n>(t); _ot<n+1>(os, t); }
template<class...T> ostream & operator<<(ostream & os, tuple<T...> const & t){ _ot<0>(os, t); return os; }
template<int n, class...T> typename enable_if<(n>=sizeof...(T))>::type _it(istream &, tuple<T...> &){}
template<int n, class...T> typename enable_if<(n< sizeof...(T))>::type _it(istream & is, tuple<T...> & t){ is >> get<n>(t); _it<n+1>(is, t); }
template<class...T> istream & operator>>(istream & is, tuple<T...> & t){ _it<0>(is, t); return is; }
template<class T, class U> istream & operator<<(istream & is, pair<T,U> & p){ return is >> p.first >> p.second; }
template<class T, class U> ostream & operator<<(ostream & os, pair<T,U> const & p){ return os << "(" << p.first << ", " << p.second << ") "; }
template<class T> istream & operator>>(istream & is, vector<T> & v){ rep(i,v.size()) is >> v[i]; return is; }
template<class T> ostream & operator<<(ostream & os, vector<T> const & v){ rep(i,v.size()) os << v[i] << (i+1==(int)v.size()?"":" "); return os; }
#ifdef DEBUG
#define dump(...) (cerr<<#__VA_ARGS__<<" = "<<mt(__VA_ARGS__)<<" ["<<__LINE__<<"]"<<endl)
#else
#define dump(...)
#endif

const int MAX_N = 10000; //400MB
// const int MAX_N = 1024; // 4MB
// nCr % mod
ll nCr(int n, int r){
    static bool done[MAX_N+1][MAX_N/2+2];
    static ll dp[MAX_N+1][MAX_N/2+2]; // 400MB
    if (n < 0 || r < 0 || n < r) return 0;
    if (r==0) return 1;
    if (r > n-r) r = n-r;
    ll &res = dp[n][r];
    if (done[n][r]) return res;
    done[n][r] = true;
    return res = (nCr(n-1,r-1)+nCr(n-1,r));
}


double P[120], T[120], V[120];

int main(){
    int N,M;
    double L;
    while(cin >> N >> M >> L){
        rep(i,N){
            cin >> P[i] >> T[i] >> V[i];
            P[i]/=100;
        }
        rep(i,N){
            double ans = 0;
            rep(j,M+1){
                double t = L/V[i]+j*T[i];
                double x = 1;
                rep(ii,N)if(i!=ii){
                    double y = 0;
                    rep(jj,M+1){
                        double tt = L/V[ii]+jj*T[ii];
                        if(t<tt && abs(t-tt)>1e-10){
                            y += nCr(M,jj)*pow(P[ii],jj)*pow(1-P[ii],M-jj);
                        }
                    }
                    x*=y;
                }
                ans += x*nCr(M,j)*pow(P[i],j)*pow(1-P[i],M-j);
            }
            printf("%.10lf\n", ans);
        }
    }
}