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;

int N, M, L;
vector<double> P, T, V;
double F[51];
double C[51][51];

int main(){
    cout.precision(16);
    cout.setf(ios::fixed);
    
    F[0] = 1;
    for(int i=1; i<51; i++)
        F[i] = F[i-1] * i;
    for(int i=0; i<51; i++)
        for(int j=0; j<=i; j++)
            C[i][j] = F[i] / F[i-j] / F[j];
    
    cin >> N >> M >> L;
    P.resize(N); T.resize(N); V.resize(N);
    for(int i=0; i<N; i++)
        cin >> P[i] >> T[i] >> V[i];
    
    for(int i=0; i<N; i++){
        double ans = 0;
        for(int k=0; k<=M; k++){ //????????????i????????????????????????
            double ti = L / V[i] + T[i]*k;
            double ttans = 1;
            for(int j=0; j<N; j++) if(i != j){
                double tans = 0;
                for(int l=0; l<=M; l++){ //????????????j????????????????????????
                    double tj = L / V[j] + T[j]*l;
                    if(ti < tj){
                        tans += pow(P[j]/100, l) * pow(1-P[j]/100, M-l) * C[M][l];
                    }
                }
                ttans *= tans;
            }
            ans += ttans * pow(P[i]/100, k) * pow(1-P[i]/100, M-k) * C[M][k];
        }
        cout << ans << endl;
    }
    
    return 0;
}