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 <algorithm>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>

using namespace std;

typedef long long LL;
const LL maxn = 111;
LL v[maxn], t[maxn];
LL c[maxn][maxn];

double p[maxn];
double lv[maxn];
LL n, m, l;

double ff (LL u, LL state, LL v) {
	double ret = 0;
	for (LL i = 0; i <= m; ++i) {
		if (lv[v] + i * t[v] > lv[u] + state * t[u]) {
			ret += 1.0 * c[i][m] * pow (1.0 * p[v], 1.0*i) * pow (1.0 - p[v], 1.0*m - i);
		}
	}
	return ret;
}
double f (LL u, LL state) {
	double ret = 1.0 * c[state][m] * pow (1.0 * p[u], 1.0 * state) * pow (1.0 - p[u], 1.0 * m - state);
	for (LL i = 0; i < n; ++i) {
		if (i == u) continue;
		ret *= ff (u, state, i);
	}
	return ret;
}
double solve (LL u) {
	double ret = 0;
	for (LL i = 0; i <= m; ++i) {
		ret += f (u, i);
	}
	return ret;
}

//快速求&#32452;合数 c(m, n)
LL combination(LL m, LL n) {
	LL i, a, b, p;
	if (n < m) n ^= m ^= n ^= m;
	p = 1;
	a = n - m < m ? n-m : m;
	b = n - m > m ? n-m : m;
	for (i = 1; i <= a; ++i) p += p * b / i;
	return p;
}

int main () {
	LL i;
	while (~scanf ("%lld%lld%lld", &n, &m, &l)) {
		//生成所有c[i][m]的&#32452;合数
		for (i = 0; i <= (m >> 1); ++i) {
			c[i][m] = c[m-i][m] = combination (i, m);
		}
		/*for (i = 0; i <= m; ++i) {
			printf ("c[%lld][%lld] = %lld\n", i, m, c[i][m]);
		}*/
		for (i = 0; i < n; ++i) {
			scanf ("%lf%lld%lld", &p[i], &t[i], &v[i]);
			lv[i] = 1.0 * l / v[i];
			p[i] /= 100;
		}
		for (i = 0; i < n; ++i) {
			double ans = solve (i);
			printf ("%.8lf\n", ans);
		}
	}
	return 0;
}