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 <iostream>
#include <vector>
#include <cmath>
#include <iomanip>
using namespace std;

constexpr double eps = 1e-10;

int main() {
    int N, M, L;
    cin >> N >> M >> L;
    vector<vector<vector<double>>> cmb(N, vector<vector<double>>(M+1, vector<double>(M+1)));
    for(int i=0; i<N; ++i) {
        cmb[i][0][0] = 1;
    }
    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;
    }
    for(int i=0; i<N; ++i) {
        for(int j=1; j<=M; ++j) {
            for(int k=0; k<=j; ++k) {
                cmb[i][j][k] = cmb[i][j-1][k] * (1 - P[i]) + (k-1 >= 0 ? cmb[i][j-1][k-1] * P[i] : 0);
            }
        }
    }
    for(int i=0; i<N; ++i) {
        if(V[i] == 0) {
            cout << 0 << endl;
            continue;
        }
        double res = 0;
        for(int j=0; j<=M; ++j) {
            double t = L / V[i] + T[i] * j;
            double p = cmb[i][M][j];
            for(int k=0; k<N; ++k) {
                if(i == k || V[k] == 0) {
                    continue;
                }
                double tt = L / V[k];
                if(T[k] == 0 && tt - t <= 0) {
                    p = 0;
                    break;
                }
                int cnt = (T[k] == 0 ? 0 : ceil(max(0.0, t - tt) / T[k]));
                if(tt + T[k] * cnt - t < eps) {
                    cnt += 1;
                } 
                double pp = 0;
                for(int l=cnt; l<=M; ++l) {
                    pp += cmb[k][M][l];
                }
                p *= pp;
            }
            res += p;
        }
        cout << fixed << setprecision(10) << res << endl;
    }
}