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 <stdio.h>
#include <cstring>

using namespace std;

const int MAXN = 110;
const double eps = 1e-12;

double pro[MAXN][55][55];
double speed[MAXN], rest[MAXN], p[MAXN];
double to[MAXN][55];
double dis;
int n, m;

void read()
{
    for(int i = 0; i < n; ++i)
    {
        scanf("%lf%lf%lf", p + i, rest + i, speed + i);
        p[i] *= 0.01;
    }

    memset(pro, 0, sizeof(pro));
    for(int i = 0; i < n; ++i)
    {
        pro[i][0][0] = 1.0;
        for(int j = 1; j <= m; ++j)
        {
            pro[i][j][0] = pro[i][j - 1][0] * (1 - p[i]);
            for(int k = 1; k <= j; ++k)
                pro[i][j][k] = pro[i][j -  1][k] * (1 - p[i]) + pro[i][j - 1][k - 1] * p[i];
        }
    }

    for(int i = 0; i < n; ++i)
    {
        for(int j = 0; j <= m; ++j)
            to[i][j] = dis / speed[i] + j * rest[i];
    }
}
double ans[MAXN];
int main()
{
    while(scanf("%d%d%lf", &n, &m, &dis) != EOF)
    {
        read();
        for(int i = 0; i < n; ++i)
        {
            ans[i] = 0.0;
            for(int j = 0; j <= m; ++j)
            {
                double hh(pro[i][m][j]);
                for(int k = 0; k < n; ++k)
                {
                    if(k == i)  continue;
                    double tmp = 1.0;
                    for(int u = 0; u <= m; ++u)
                    {
                        if(to[k][u] > to[i][j] + eps)  break;
                        tmp -= pro[k][m][u];
                    }
                    hh *= tmp;
                }
                ans[i] += hh;
            }
        }
        for(int i = 0; i < n; ++i)
            printf("%lf\n", ans[i]);
    }


    return 0;
}