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:
// OBツ嘉ッツ嘉債宿2011 Day3 D : Marathon Match

#include <iostream>
#include <queue>
#include <cstdio>
#include <cstring>

using namespace std;

int main(){
	int N, M, L;
	int P[100], T[100], V[100];
	// [P][M][lest]
	static double dp[101][101][101];
	for(int i=0;i<=100;i++){
		for(int j=0;j<=100;j++)
			for(int k=0;k<=100;k++) dp[i][j][k] = 0.0;
		dp[i][0][0] = 1.0;
		for(int j=0;j<100;j++){
			for(int k=0;k<=j;k++){
				dp[i][j+1][k  ] += dp[i][j][k]*(100-i)/100.0;
				dp[i][j+1][k+1] += dp[i][j][k]*i/100.0;
			}
		}
	}
	while(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++){
			double res = 0.0;
			for(int j=0;j<=M;j++){
				double cur = dp[P[i]][M][j];
				if(V[i]==0) break;
				double time = (double)L/V[i] + j*T[i];
				for(int k=0;k<N;k++){
					if(i==k) continue;
					if(V[k]==0) continue;
					double sum = 0.0;
					for(int l=0;l<=M;l++){
						double otime = (double)L/V[k] + l*T[k];
						if(otime - time < 1e-8) continue;
						sum += dp[P[k]][M][l];
					}
					cur *= sum;
				}
				res += cur;
			}
			printf("%.8lf\n", res);
		}
	}
}