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 <math.h>
#define rep(i, n) for(int i=0; i<(int)(n); i++)
typedef long long Int;

int N, M, L, P[200], T[200], V[200];
double zp[200][200], zt[200][200];

Int C[100][100];

int main() {
    C[0][0] = 1;
    rep(i, 80) rep(j, 80) C[i+1][j]+=C[i][j], C[i+1][j+1]+=C[i][j];

    scanf("%d%d%d", &N, &M, &L);
    rep(i, N) scanf("%d%d%d", P+i, T+i, V+i);
    rep(i, N) rep(j, M+1) {
        const double p = P[i]/100.0;
        zt[i][j] = (double)L/V[i] + (double)j*T[i];
        zp[i][j] = (double)C[M][j] * pow(p, j) * pow(1-p, M-j);
    }
    rep(i, N) {
        double ans = 0;
        if(V[i]!=0) rep(j, M+1) {
            double f = 1.0;
            rep(k, N) if(i!=k) {
                double p = 0;
                rep(l, M+1) if(zt[i][j]<zt[k][l]) p += zp[k][l];
                f *= p;
            }
            ans += zp[i][j] * f;
        }
        printf("%.9f\n", ans);
    }
    return 0;
}