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;

#define MAX_N 100
#define MAX_M 50
#define INF (1<<29)

int main()
{
    int N, M;
    double L;
    cin >> N >> M >> L;
    vector<double> P(N), T(N), V(N);
    for (int i = 0; i < N; i++) {
        cin >> P[i] >> T[i] >> V[i];
        P[i] /= 100;
    }
    double p[MAX_N][MAX_M+1][MAX_M+1];
    memset(p, 0, sizeof(p));

    for (int i = 0; i < N; i++) {
        p[i][0][0] = 1;
        for (int j = 0; j < M; j++) {
            for (int k = 0; k < M; k++) {
                p[i][j+1][k]   += p[i][j][k] * (1 - P[i]);
                p[i][j+1][k+1] += p[i][j][k] * P[i];
            }
        }
    }
    
    for (int i = 0; i < N; i++) {
        double res = 0;
        for (int j = 0; j <= M; j++) {
            double r = 1;
            bool win = 1;
            for (int k = 0; k < N; k++) {
                if (i == k) {
                    continue;
                }
                double rr = 0;
                bool ok = 0;
                for (int l = M; l >= 0; l--) {
                    if (V[k]*L+V[k]*V[i]*T[i]*j <
                        V[i]*L+V[i]*V[k]*T[k]*l) {
                        ok = 1;
                        rr += p[k][M][l];
                    } else {
                        break;
                    }
                }
                if (!ok) {
                    win = 0;
                    break;
                }
                r *= rr;
            }
            res += (win ? r * p[i][M][j] : 0);
        }
        printf("%.8f\n", res);
    }
    return 0;
}