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 vector<int> vi;
typedef vector<double> vd;

double mypow(double b ,int e){
	  if(e == 0) return 1;
	  if(e%2==0)
			return mypow( b*b, e/2 );
	  else
			return mypow( b*b, e/2 ) * b;
}

int main(){


	  int N,M,L; cin >> N >> M >> L ;
	  if(N==1){
			cout << 1.0 << endl;
			return 0;
	  }
	  vector<vd> prob(N,vd(M+1,0.0));
	  vector<vd> dist(N,vd(M+1,0.0));

	  vd p(N),t(N),v(N);
	  for(int i=0; i<N; i++)	cin >> p[i] >> t[i] >> v[i];

	  // PRE PROCESS
	  for(int i=0; i<N; i++){
			dist[i][0] = (double)L / v[i];
			for(int j=1; j<=M; j++){
				  dist[i][j] = dist[i][j-1] + t[i];
			}
	  }
	  for(int i=0; i<N; i++)
			prob[i][0] = mypow((100-p[i])/100,M);
	  for(int i=0; i<N; i++){
			for(int j=1; j<=M; j++){
				  if(p[i] != 100)
						prob[i][j] = (prob[i][j-1] * (p[i]/100)) / ((100-p[i])/100);
				  else
						prob[i][j] = 1.0;
			}
	  }
	  double comb = 1;
	  for(int j=1; j<=M; j++){
			comb = comb * (M - j + 1) / j; 
			for(int i=0; i<N; i++){
				  prob[i][j] *= comb;
			}
	  }
	  
	  vector<vd> sum = prob;
	  for(int i=0; i<N; i++){
			for(int j=M-1; j>=0; j--){
				  sum[i][j] += sum[i][j+1];
				  if(sum[i][j] > 1) sum[i][j] = 1;
			}
	  }

	  vector<vd> ans = prob;
	  for(int i=0; i<N; i++){
			for(int j=0; j<=M; j++){
				  for(int k=0; k<N; k++){
						if(i==k) continue;
						int id = upper_bound(dist[k].begin(),dist[k].end(), dist[i][j]) - dist[k].begin();
						if(id > M) ans[i][j] = 0.0;
						else 	   ans[i][j] *= sum[k][id];

				  }
			}
	  }

	  for(int i=0; i<N; i++){
			double tmp = 0.0;
			if(p[i] == 100) tmp = ans[i][M];
			else
				  for(int j=0; j<=M; j++)
						tmp += ans[i][j];
			if(tmp > 1) tmp = 0;
			printf("%.10f\n",tmp);
	  }
						
}