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;
typedef long long ll;
int P[100], T[100], V[100];
double prob[100][51];
double comb[51][51];
int main() {
	int N, M, L;
	cin >> N >> M >> L;
	comb[0][0] = 1;
	for (int i = 1; i <= M; i++) {
		comb[i][0] = comb[i][i] = 1;
		for (int j = 1; j < i; j++) {
			comb[i][j] = comb[i - 1][j - 1] + comb[i - 1][j];
		}
	}
	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++) {
			prob[i][j] = comb[M][j] * pow(P[i] / 100.0, j)*pow(1 - P[i] / 100.0, M - j);
		}
	}
	for (int i = 0; i < N; i++) {
		if (V[i] == 0) {
			cout << 0 << endl;
			continue;
		}
		double ans = 0;
		for (int j = 0; j <= M; j++) {
			double t1 = (double)L / V[i] + T[i] * j;
			double p = 1;
			for (int k = 0; k < N; k++) {
				if (k == i) continue;
				if (V[k] == 0) continue;
				double sum = 0;
				for (int l = 0; l <= M; l++) {
					double t2 = (double)L / V[k] + T[k] * l;
					if (t1 < t2) sum += prob[k][l];
				}
				p *= sum;
			}
			ans += prob[i][j] * p;
		}
		printf("%.15lf\n", ans);
	}
}