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 <iomanip>
#include <iostream>
using namespace std;
int n, m, l, p[111], t[111], v[111]; long double r[111][55], dp[55][55];
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++) {
		for (int j = 0; j <= m; j++) fill(dp[j], dp[j] + m + 1, 0.0);
		dp[0][0] = 1;
		for (int j = 0; j < m; j++) {
			for (int k = 0; k <= j; k++) {
				dp[j + 1][k] += dp[j][k] * (100 - p[i]) / 100;
				dp[j + 1][k + 1] += dp[j][k] * p[i] / 100;
			}
		}
		for (int j = 0; j <= m; j++) r[i][j] = dp[m][j];
	}
	for (int i = 0; i < n; i++) {
		long double ret = 0;
		if (v[i] != 0) {
			long double time1 = 1.0L * l / v[i];
			for (int j = 0; j <= m; j++, time1 += t[i]) {
				long double res = r[i][j];
				for (int k = 0; k < n; k++) {
					if (i == k) continue;
					long double sum = 0;
					if (v[k] != 0) {
						double time2 = 1.0 * l / v[k];
						for (int l = 0; l <= m; l++, time2 += t[k]) {
							if (time1 + 1e-10L < time2) sum += r[k][l];
						}
					}
					else sum = 1;
					res *= sum;
				}
				ret += res;
			}
		}
		cout << fixed << setprecision(15) << ret << endl;
	}
	return 0;
}