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 <algorithm>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <vector>
using namespace std;

const int MAX_M = 50;
long long memo[MAX_M + 1][MAX_M + 1];
long long combination(int n, int k) {
	if(k == 0 || k == n)
		return 1;

	if(k * 2 > n)
		k = n - k;

	if(memo[n][k] > 0)
		return memo[n][k];

	return memo[n][k] = combination(n - 1, k - 1) + combination(n - 1, k);
}

int main() {
	int n, m, l;
	scanf("%d %d %d", &n, &m, &l);

	vector<int> t(n), v(n);
	vector<double> p(n);
	for(int i = 0; i < n; ++i) {
		scanf("%lf %d %d", &p[i], &t[i], &v[i]);
		p[i] /= 100.0;
	}

	vector<vector<double>> time(n, vector<double>(m + 1));
	for(int i = 0; i < n; ++i) {
		time[i][0] = static_cast<double>(l) / v[i];
		for(int j = 1; j <= m; ++j) {
			time[i][j] = time[i][j - 1] + t[i];
		}
	}

	vector<vector<double>> probability(n, vector<double>(m + 1));
	for(int i = 0; i < n; ++i) {
		const double q = 1.0 - p[i];
		for(int j = 0; j <= m; ++j) {
			probability[i][j] = combination(m, j) * pow(p[i], j) * pow(q, m - j);
		}
	}

	vector<vector<double>> sum(probability);
	for(int i = 0; i < n; ++i) {
		for(int j = m - 1; j >= 0; --j) {
			sum[i][j] += sum[i][j + 1];
		}
	}

	for(int i = 0; i < n; ++i) {
		double ans = 0.0;
		for(int j = 0; j <= m; ++j) {
			double tmp = probability[i][j];
			for(int k = 0; k < n; ++k) {
				if(i == k) continue;

				const int idx = upper_bound(time[k].begin(), time[k].end(), time[i][j]) - time[k].begin();
				tmp *= sum[k][idx];
			}
			ans += tmp;
		}

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

	return EXIT_SUCCESS;
}