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:
#include "bits/stdc++.h"
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
const int INF = 1e9;
const ll LINF = 1e18;
template<class S,class T> ostream& operator << (ostream& out,const pair<S,T>& o){ out << "(" << o.first << "," << o.second << ")"; return out; }
template<class T> ostream& operator << (ostream& out,const vector<T> V){ for(int i = 0; i < V.size(); i++){ out << V[i]; if(i!=V.size()-1) out << " ";} return out; }
template<class T> ostream& operator << (ostream& out,const vector<vector<T> > Mat){ for(int i = 0; i < Mat.size(); i++) { if(i != 0) out << endl; out << Mat[i];} return out; }
template<class S,class T> ostream& operator << (ostream& out,const map<S,T> mp){ out << "{ "; for(auto it = mp.begin(); it != mp.end(); it++){ out << it->first << ":" << it->second; if(mp.size()-1 != distance(mp.begin(),it)) out << ", "; } out << " }"; return out; }

/*
 <url:http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=2303>
 問題文============================================================
 =================================================================
 解説=============================================================
 ================================================================
 */

double powmod(double x, ll n) {
    double res = 1;
    while (n > 0) {
        if (n & 1) { //奇数なら
            res =(res*x);
        }
        x = (x*x);
        n >>= 1;
    }
    return res;
}
void solve(){
    int N,M,L; cin >> N >> M >> L;
    vector<double> P(N),T(N),V(N);
    for(int i = 0; i < N;i++){
        cin >> P[i] >> T[i] >> V[i];
        P[i]/=100.;
    }
    const int MAX_M = 50;
    double comb[MAX_M+1][MAX_M+1];
    for (int i = 0; i <= MAX_M; i++) {
        for (int j = 0; j <= i; j++) {
            if (j == 0 || j == i)
                comb[i][j] = 1;
            else
                comb[i][j] = comb[i - 1][j - 1] + comb[i - 1][j];
        }
    }
    
    cout << fixed << setprecision(12);
    for(int i = 0; i < N;i++){
        if(V[i] == 0){ cout << 0. << endl; continue;}
        double ans = 0;
        for(int j = 0; j <= M;j++){
            double tans = powmod(P[i],j)*powmod(1-P[i],M-j)*comb[M][j];
            double Time = T[i]*j + L/V[i];
            for(int k = 0; k < N;k++){
                if(i == k) continue;
                double sum = 0;
                for(int l = 0; l <= M; l++){
                    double TTime = T[k]*l + L/V[k];
                    if(TTime < Time + 1e-9) continue;
                    sum += powmod(P[k],l)*powmod(1-P[k],M-l)*comb[M][l];
                }
                tans *= sum;
            }
            ans += tans;
        }
        cout << ans << endl;
    }
}
int main(void) {
    cin.tie(0); ios_base::sync_with_stdio(false);
    solve();
    return 0;
}