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;
typedef long long LL;
const int MAXN = 55;
double C[MAXN+1][MAXN+1];
const int N = 110;
int n, m;
int l, t[N], v[N];
double p[N];
void init()
{
    memset(C, 0, sizeof C);
    for(int i=0; i<=MAXN; ++i)
    {
        C[0][i] = 0.0;
        C[i][0] = 1.0;
    }
    for(int i=1; i<=MAXN; ++i)
    {
        for(int j=1; j<=MAXN; ++j)
        C[i][j] = C[i-1][j] + C[i-1][j-1];
    }
}
double work(int x)
{
    double ans = 0.0;
    for(int i = 0; i <= m; i++)
    {
        double pp = C[m][i] * pow(p[x], i) * pow(1.0-p[x], m-i);
        for(int j = 1; j <= n; j++)
        {
            if(j == x)continue;
            double siz = 0;
            for(int k = m; k >= 0; k--)
            {
                if(l*v[j] + i*t[x]*v[j]*v[x] < l*v[x] + k*t[j]*v[x]*v[j])
                    siz += C[m][k] * pow(p[j], k) * pow(1.0-p[j], m-k);
                else break;
            }
            pp *= siz;
        }
        ans += pp;
    }
    return ans;
}
int main()
{
    //freopen("in.txt","r",stdin);
    init();
    while(cin>>n>>m>>l)
    {
        for(int i = 1; i <= n; i++)
        {
            cin>>p[i]>>t[i]>>v[i];
            p[i] /= 100.0;
        }
        for(int i = 1; i <= n; i++)
            printf("%.10f\n", work(i));
    }
    return 0;
}