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 <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cmath>

using namespace std;

const int MAXN = 100 + 10;
const int MAXM = 6000 + 10;
const double eps = 1e-8;

int n, m, l, tot;
double org[MAXN];
double p[MAXN], t[MAXN], v[MAXN];
double b[MAXM], c[MAXM], tp1[MAXM], tp2[MAXM];
double C[MAXN][MAXN];

int Find(double x) {
        for (int i = 1; i <= tot; ++i)
		if (fabs(x - c[i]) < eps)
			return i;
}

double co[MAXN][MAXM];

inline int lowbit(int x) {
	return x & (-x);
}

inline void Add(int i, int x, double v) {
	if (x == 0)
		return;
	for (; x <= tot; x += lowbit(x))
		co[i][x] += v;
}

inline double Query(int i, int x) {
	double ret = 0;
	for (; x > 0; x -= lowbit(x))
		ret += co[i][x];
	return ret;
}
int main() {
	C[0][0] = 1;
	for (int i = 1; i <= 50; ++i) {
		C[i][0] = 1;
		for (int j = 1; j <= i; ++j)
			C[i][j] = C[i - 1][j - 1] + C[i - 1][j];
	}
	scanf("%d%d%d", &n, &m, &l);
	for (int i = 1; i <= n; ++i) {
		scanf("%lf%lf%lf", &p[i], &t[i], &v[i]);
		org[i] = 1.0 * l / v[i];
		p[i] = p[i] / 100;
	}
	for (int i = 1; i <= n; ++i)
		for (int j = 0; j <= m; ++j) {
			b[++tot] = -(org[i] + j * t[i]);
		}
	sort(b + 1, b + 1 + tot);
	int tmp = 0;
	for (int i = 1; i <= tot; ++i) {
		if (i == 1 || fabs(b[i] - b[i - 1]) > eps) {
			c[++tmp] = b[i];
		}
	}
	tot = tmp;
	for (int i = 1; i <= n; ++i) {
		for (int j = 0; j <= m; ++j) {
			double curs = -(org[i] + j * t[i]);
			double curp = C[m][j];
		        for (int k = 1; k <= j; ++k)
				curp *= p[i];
			for (int k = 1; k <= m - j; ++k)
				curp *= (1 - p[i]);
			int cur = Find(curs);
			Add(i, cur, curp);
		}
	}
	for (int i = 1; i <= n; ++i) {
		double ans = 0;
		for (int j = 0; j <= m; ++j) {
			double curs = -(org[i] + j * t[i]);
			double curp = C[m][j];
			for (int k = 1; k <= j; ++k)
				curp *= p[i];
			for (int k = 1; k <= m - j; ++k)
				curp *= (1 - p[i]);
			//	cout << i << ' ' << j << ' ' << curp << endl;
			int cur = Find(curs);
			for (int k = 1; k <= n; ++k) if (i != k) {
					curp *= Query(k, cur - 1);
			}
			ans += curp;
		}
		printf("%.8f\n", ans);
	}

	return 0;
}