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;
const int maxn = 101,maxm = 51;
int P[maxn],T[maxn],V[maxn];
long long C[maxm][maxm];
double rst[maxn][maxm];
double tim[maxn][maxm];
const double eps = 1e-11;
//#define LOCAL
int main()
{
#ifdef LOCAL
    freopen("in.txt","r",stdin);
#endif
    int n,m,l; scanf("%d%d%d",&n,&m,&l);
    C[0][0] = 1;
    for(int i = 1; i < maxm; i++){
        C[i][i] = C[i][0] = 1;
        for(int j = 1; j < i; j++){
            C[i][j] = C[i-1][j] + C[i-1][j-1];
        }
        //cout<<C[i][i/2]<<endl;
    }
    for(int i = 0; i < n; i++){
        scanf("%d%d%d",P+i,T+i,V+i);
    }
    int rt = 0;
    for(int i = 0; i < n; i++){
        double p = P[i]/100., t0 = l*1./V[i];
        for(int j = 0; j <= m; j++){
            rst[i][j] = C[m][j]*(pow(p,j))*(pow(1-p,m-j));
            tim[i][j] = t0+j*T[i];
        }
        //for(int j = 1; j <= m; j++){
       //     rst[i][j] += rst[i][j-1];
      //  }
    }
    for(int i = 0; i < n; i++){
        double ans = 0;
        for(int j = 0; j <= m; j++){
            double t = tim[i][j], wn = 1;
            for(int k = 0; k < n; k++){
                if(k == i) continue;
                double twn = 0;
                for(int j2 = m; j2 >= 0; j2--){
                    if(tim[k][j2] <= t) break;
                    twn += rst[k][j2];
                }
                wn *= twn;
            }
            ans += wn*rst[i][j];
        }
        printf("%lf\n",ans);
    }
    return 0;
}