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<sstream>
#include<algorithm>
#include<set>
#include<map>
#include<queue>
#include<complex>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cassert>

#define rep(i,n) for(int i=0;i<(int)n;i++)
#define all(c) (c).begin(),(c).end()
#define mp make_pair
#define pb push_back
#define each(i,c) for(__typeof((c).begin()) i=(c).begin();i!=(c).end();i++)
#define dbg(x) cerr<<__LINE__<<": "<<#x<<" = "<<(x)<<endl

using namespace std;

typedef long long ll;
typedef vector<int> vi;
typedef pair<int,int> pi;
const int inf = (int)1e9;
const double INF = 1e12, EPS = 1e-9;

int n, m, l, p[100], t[100], v[100];
double dp[100][52], sum[100][52];

int main(){
	cin >> n >> m >> l;
	rep(i, n) cin >> p[i] >> t[i] >> v[i];
	rep(i, n){
		dp[i][0] = 1;
		rep(j, m) for(int k = m; k >= 0; k--){
			dp[i][k + 1] += dp[i][k] * p[i] / 100.0;
			dp[i][k] *= 1 - p[i] / 100.0;
		}
		sum[i][m] = dp[i][m];
		for(int j = m - 1; j >= 0; j--) sum[i][j] = sum[i][j + 1] + dp[i][j];
		/*
		rep(j, m + 1) cerr<<dp[i][j]<<(j==m?"\n":" ");
		rep(j, m + 1) cerr<<sum[i][j]<<(j==m?"\n":" ");
		*/
	}
	
	rep(i, n){
		double s = 0;
		if(v[i]) rep(j, m + 1){
			double e = dp[i][j];
			rep(k, n) if(k != i){
				int tm = (int)ceil((l * 1.0 / v[i] - l * 1.0 / v[k] + t[i] * j) / t[k] + EPS);
				
				tm = max(tm, 0);
				if(t[k] == 0) tm = l * 1.0 / v[i] - l * 1.0 / v[k] + t[i] * j < - EPS ? 0 : m + 1;
				if(v[k] == 0) tm = 0;
				if(tm <= m) e *= sum[k][tm];
				else e = 0;
			}
			s += e;
		}
		printf("%.9f\n", s);
	}
	
	return 0;
}