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 <cstring>
#include <cstdio>
using namespace std;

#define INF 999999999

int N, M, L;
double P[102], T[102], V[102];
double dp[102][102];
double p[102][102];

int main(){
  while(cin >> N >> M >> L){
    for(int i = 0; i < N; i++){
      cin >> P[i] >> T[i] >> V[i];
      P[i] /= 100.0;
    }

    for(int k = 0; k < N; k++){
      memset(dp, 0, sizeof(dp));
      dp[0][0] = 1.0;

      for(int i = 0; i < M; i++){
        for(int j = 0; j <= M; j++){
          dp[i + 1][j] += dp[i][j] * (1 - P[k]);
          dp[i + 1][j + 1] += dp[i][j] * P[k];
        }
      }

      for(int j = 0; j <= M; j++){
        p[k][j] = dp[M][j];
      }
    }

    for(int i = 0; i < N; i++){
      double ans = 0;

      for(int j = 0; j <= M; j++){
        double ti = INF;
        if(V[i] != 0) ti = (double)j * T[i] + (double)L / V[i];

        double mul = 1.0;

        for(int k = 0; k < N; k++){
          if(k == i) continue;

          double sum = 0;

          for(int l = 0; l <= M; l++){
            double tk = INF;
            if(V[k] != 0) tk = (double)l * T[k] + (double)L / V[k];
            if(ti >= tk) continue;
            sum += p[k][l];
          }

          mul *= sum;
        }

        ans += mul * p[i][j];
      }

      printf("%.8lf\n", ans);
    }
  }
}