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;

int main() {
    int N, M, 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];
    for (int i = 0; i < N; i++)
        P[i] = P[i] / 100.;

    vector<vector<double> > times(N, vector<double>(M + 1)), prob(N, vector<double>(M + 1));
    for (int i = 0; i < N; i++)
        times[i][0] = (double)L / V[i];

    vector<vector<long long int> > comb(M + 1, vector<long long int>(M + 1));
    for (auto& i : comb) i[0] = 1LL;
    for (int i = 1; i <= M; i++)
        for (int j = 1; j <= M; j++) {
            long long int up = comb[i-1][j];
            long long int left = comb[i-1][j-1];
            comb[i][j] = up + left;
        }

    for (int i = 0; i < N; i++) {
        for (int j = 0; j <= M; j++) {
            if (j > 0) times[i][j] = times[i][j-1] + T[i];
            prob[i][j] = pow(P[i], j) * pow(1.000 - P[i], M - j) * comb[M][j];
        }
    }

    for (int i = 0; i < N; i++) {
        double ans = 0.0000;
        for (int j = 0; j <= M; j++) {
            double tmp = 1.0000;
            for (int k = 0; k < N; k++) {
                if (k == i) continue;
                double sum = 0.0000;
                for (int l = 0; l <= M; l++) {
                    if (times[i][j] < times[k][l]) sum += prob[k][l];
                }
                tmp *= sum;
            }
            ans += prob[i][j] * tmp;
        }
        printf("%.10f\n", ans);
    }


    return 0;
}