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(){
  cout << fixed << setprecision(5);
  int N, M, L;
  cin >> N >> M >> L;
  vector<int> P(N), T(N), V(N);
  for (int i = 0; i < N; i++){
    cin >> P[i] >> T[i] >> V[i];
  }
  vector<vector<vector<double>>> dp(N, vector<vector<double>>(M + 1, vector<double>(M + 1, 0)));
  for (int i = 0; i < N; i++){
    dp[i][0][0] = 1;
    double p = (double) P[i] / 100;
    for (int j = 0; j < M; j++){
      for (int k = 0; k <= j; k++){
        dp[i][j + 1][k] += dp[i][j][k] * (1 - p);
        dp[i][j + 1][k + 1] += dp[i][j][k] * p;
      }
    }
  }
  for (int i = 0; i < N; i++){
    double ans = 0;
    for (int j = 0; j <= M; j++){
      double t = (double) L / V[i] + j * T[i];
      double tmp = dp[i][M][j];
      for (int k = 0; k < N; k++){
        if (k != i){
          double sum = 0;
          for (int l = 0; l <= M; l++){
            if ((double) L / V[k] + l * T[k] <= t){
              sum += dp[k][M][l];
            }
          }
          tmp *= 1 - sum;
        }
      }
      ans += tmp;
    }
    cout << ans << endl;
  }
}