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 <iostream>
#include <string>
#include <cstdio>
#include <vector>
#include <cstring>
#include <map>
#include <algorithm>
using namespace std;

#define FOR(i,c) for (__typeof((c).begin())i=(c).begin(); i!=(c).end(); ++i)

int main() {
    int n, m , L;
    cin >> n >> m >> L;
    int v[n];
    double t[n][m+1], p[n][m+1], sump[n][m+1];
    for (int i=0; i<n; ++i) {
        double pp, tt;
        cin >> pp>>tt>>v[i];
        pp/=100;
        
        for (int k=0; k<=m; ++k) {
            p[i][k] = 1;
            for (int l=0; l<=m; ++l) {
                if (l < k) p[i][k] *= m-l;
                if (k-l >= 1) {
                    p[i][k] /= k-l;
                }
                if (l<k) p[i][k] *= pp;
                if (l<m-k) p[i][k] *= (1.0-pp);
            }
            t[i][k] = (double)L/v[i] + k*tt;
//            printf("i=%d, k=%d %f\n", i,k,p[i][k]);
            if (k>0)
                sump[i][k] = sump[i][k-1] + p[i][k];
            else
                sump[i][k] = p[i][k];
        }
    }
    
    double ans[n];
    for (int i=0;i<n;++i) ans[i] = 0;

    for (int i=0; i<n; ++i) {
        for (int j=0; j<=m; ++j) {           
            double tmp = p[i][j];
            for (int k=0; k<n; ++k) {
                if (i==k) continue;
                if (upper_bound(t[k], t[k]+m+1, t[i][j])-t[k] == 0) {
                    
//                    printf("zero %d %d\n", i, j);
                    // tmp = 0;
                    // break;
                }
                else 
                    tmp *= (1.0 - sump[k][upper_bound(t[k], t[k]+m+1, t[i][j])-t[k]-1]);
            }
//            printf("%d,%d %f\n", i,j,tmp);
            ans[i] += tmp;
        }
    }

    for (int i=0;i<n;++i)
        printf("%.12f\n", ans[i]);
}