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<iomanip>
using namespace std;

#define MAX_N 1000

double x[MAX_N][MAX_N];
double t[MAX_N][MAX_N];
int n, m, l;
int p[MAX_N][3];
double dp[MAX_N][MAX_N];
double boss[MAX_N];
double kakuritsu;
double res;

int main() {
	cin >> n >> m >> l;
	for (int h = 0; h < n; h++) {
		cin >> p[h][0] >> p[h][1] >> p[h][2];
		//kakuritsu calc.
		memset(dp, 0, sizeof(dp));
		dp[0][0] = 1.0;
		for (int i = 1; i <= m; i++) {
			for (int j = 0; j <= m; j++) {
				dp[i][j] += dp[i - 1][j] * (100.0 - p[h][0]) / 100.0;
				dp[i][j + 1] += 1.0 * dp[i - 1][j] * p[h][0] / 100.0;
			}
		}
		for (int i = 0; i <= m; i++) {
			x[h][i] = dp[m][i];
		}
		//time calc.
		for (int i = 0; i <= m; i++) {
			t[h][i] = (1.0*l / p[h][2]) + p[h][1] * i;
		}
	}
	//kakuritsu calc 2.
	for (int h = 0; h < n; h++) {
		res = 0.0;
		for (int i = 0; i <= m; i++) {
		memset(boss, 0, sizeof(boss));
		kakuritsu = 1.0;
			for (int j = 0; j < n; j++) {
				if (h != j) {
					for (int k = 0; k <= m; k++) {
						if (t[h][i] < t[j][k]) {
							boss[j] += x[j][k];
						}
					}
					kakuritsu *= boss[j];
				}
			}
			res += kakuritsu*x[h][i];
		}
		cout << fixed << setprecision(15) << res << endl;
	}
	return 0;
}