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 N, M, L;
int P[100], T[100], V[100];
double dp[101][52];
double tim[101][52];
double small[52][52];

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

  
  for(int i = 0; i < N; i++) {
    memset(small, 0, sizeof(small));
    small[0][0] = 1.0;
    for(int j = 0; j < M; j++) {
      for(int k = 0; k <= j; k++) {
        small[j + 1][k + 1] += small[j][k] / 100.0 * P[i];
        small[j + 1][k]     += small[j][k] / 100.0 * (100 - P[i]);
      }
    }
    for(int j = 0; j <= M; j++) {
      tim[i][j] = (V[i] == 0 ? 1e18 : 1.0 * L / V[i] + j * T[i]);
      dp[i][j] = small[M][j];
    }
  }
  cout << fixed << setprecision(10);
  
  for(int i = 0; i < N; i++) {
    double ret = 0;
    for(int j = 0; j <= M; j++) {
      double per = dp[i][j];
      for(int k = 0; k < N; k++) {
        if(i == k) continue;
        double qur = 0.0;
        for(int l = 0; l <= M; l++) {
          if(tim[i][j] < tim[k][l]) qur += dp[k][l];
        }
        per *= qur;
      }
      ret += per;
    }
    cout << ret << endl;
  }
}