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:
// 誤差ゲー
#include<iostream>
#include<vector>
#include<iomanip>
using namespace std;

typedef long double ld;
const ld EPS = 1e-10;

ld comb[100][51][51] = {};

int main(){
    int n, m, l;
    cin >> n >> m >> l;
    vector<ld> p(n), t(n), v(n);
    for(int i = 0; i < n; i++)  cin >> p[i] >> t[i] >> v[i], p[i]/=100.0;
    for(int i = 0; i < n; i++){
        comb[i][0][0] = 1;
        for(int j = 1; j <= m; j++){
            for(int k = 0; k <= j; k++){
                comb[i][j][k] = (k>0 ? comb[i][j-1][k-1] : 0)*p[i]+comb[i][j-1][k]*(1-p[i]);
            }
        }
    }
    for(int i = 0; i < n; i++){
        if(v[i] == 0){
            cout << 0 << endl;
            continue;
        }
        ld ans = 0;
        for(int j = 0; j <= m; j++){
            ld me = l/v[i]+t[i]*j;
            ld add = comb[i][m][j];
            for(int k = 0; k < n; k++){
                if(i == k || v[k] == 0)  continue;
                ld oppo = l/v[k]+t[k]*m;
                ld tim = 0;
                int now = m;
                while(now>=0 && oppo-me > EPS){
                    tim += comb[k][m][now--], oppo -= t[k];
                }
                add *= tim;
            }
            ans += add;
        }
        cout << fixed << setprecision(12) << ans << endl;
    }
    return 0;
}