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 <cstdio>
#include <cmath>
#include <vector>
#include <map>
using namespace std;

typedef long long lli;

const double inf = 1.0e100;
const double eps = 1e-6;
const int MAXN = 102;
const int MAXM = 52;
int N, M;
double L;
double P[MAXN];
double T[MAXN];
double V[MAXN];
lli comb[MAXM][MAXM];

vector<double> U[MAXN];
vector<double> Q[MAXN];

bool equals(double a, double b) {
  return abs(a-b) < eps;
}

void make() {
  comb[0][0] = 1;
  for(int i = 1; i < MAXM; ++i) {
    for(int j = 0; j <= i; ++j) {
      comb[i][j] = 0;
      if(j) comb[i][j] += comb[i-1][j-1];
      if(j <= i-1) comb[i][j] += comb[i-1][j];
    }
  }
}

int main() {
  make();
  while(cin >> N >> M >> L) {
    for(int i = 0; i < N; ++i) {
      cin >> P[i] >> T[i] >> V[i];
      P[i] /= 100.0;
    }

    for(int i = 0; i < MAXN; ++i) {
      U[i].clear();
      Q[i].clear();
    }
    for(int i = 0; i < N; ++i) {
      for(int k = 0; k <= M; ++k) { // rest k times.
	if(V[i] == 0.0) U[i].push_back(inf);
	else U[i].push_back(L/V[i] + T[i]*k);
	Q[i].push_back(pow(P[i],k)*pow(1.0-P[i],M-k)*comb[M][k]);
      }
    }

    for(int i = 0; i < N; ++i) {
      double res = 0.0;
      for(int k = 0; k <= M; ++k) {
	if(V[i] == 0.0) break;
	double tmp = Q[i][k];
	for(int j = 0; j < N; ++j) {
	  if(i == j) continue;
	  /*
	  vector<double>::iterator it = 
	    lower_bound(U[j].begin(), U[j].end(), U[i][k]);
	  */
	  double a = 0;
	  for(int b = 0; b < U[j].size(); ++b) {
	    if(equals(U[j][b], U[i][k]) || U[j][b] < U[i][k]) continue;
	    a += Q[j][b];
	  }
	  tmp *= a;
	}
	res += tmp;
      }
      printf("%.8f\n", res);
    }
    //puts("");
    return 0;
  }
  return 0;
}