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:

/* monkukui 競技プログラミング用のテンプレート  (ここから) */
#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <deque>
#include <list>
#include <queue>
#include <stack>
#include <vector>
#include <utility>
#include <algorithm>
#include <map>
#include <set>
#include <complex>
#include <cmath>
#include <limits>
#include <climits>
#include <ctime>
#include <cassert>
#include <numeric>
#include <functional>
#include <bitset>

using namespace std;
using lint = long long int;
using ll = long long int;
using lnt = long long int;
using graph = vector<vector<int>>;
using wgraph = vector<vector<pair<int, long long int>>>;
long long int INF = 1001001001001001LL;
int inf = 100100100;
long long int MOD = 1000000007LL;
double PI = 3.1415926535897932;
long long int di[] = {-1, 0, 1, 0, -1, 1, 1, -1};
long long int dj[] = {0, 1, 0, -1, 1, 1, -1, -1};

template<typename T1,typename T2> inline void chmin(T1 &a,T2 b){if(a>b) a=b;}
template<typename T1,typename T2> inline void chmax(T1 &a,T2 b){if(a<b) a=b;}
inline void yes(){ cout << "yes" << endl; }
inline void Yes(){ cout << "Yes" << endl; }
inline void YES(){ cout << "YES" << endl; }
inline void no(){ cout << "no" << endl; }
inline void No(){ cout << "No" << endl; }
inline void NO(){ cout << "NO" << endl; }
inline void possible(){ cout << "possible" << endl; }
inline void Possible(){ cout << "Possible" << endl; }
inline void POSSIBLE(){ cout << "POSSIBLE" << endl; }
inline void impossible(){ cout << "impossible" << endl; }
inline void Impossible(){ cout << "Impossible" << endl; }
inline void IMPOSSIBLE(){ cout << "IMPOSSIBLE" << endl; }

#define rep(i,n) for(int i = 0; i < (n); i++)
#define rrep(i,n) for(int i = 1; i <= (n); i++)
#define drep(i,n) for(int i = (n)-1; i >= 0; i--)
#define srep(i,s,t) for(int i = s; i < t; i++)
#define all(a) a.begin(),a.end()
#define rall(a) a.rbegin(),a.rend()
#define pb push_back

/* monkukui 競技プログラミング用のテンプレート (ここまで)*/

double comb(int n, int r) {
  double ue = 1;
  double sita = 1;
  for(int i = 0; i < r; i++) ue *= (n - i);
  for(int i = r; i >= 1; i--) sita *= i;
  return ue / sita;
}

int main(){
  
  double l;
  int n, m; cin >> n >> m >> l;
    
  vector<double> p(n);
  vector<double> t(n);
  vector<double> v(n);
  
  for(int i = 0; i < n; i++) cin >> p[i] >> t[i] >> v[i];

  vector<vector<double>> g(n, vector<double> (m + 1, 0.0));

  for(int i = 0; i < n; i++) {
    for(int j = 0; j <= m; j++) {
      // i 番目の人が, j 回休む確率
      double val = pow(p[i] / (double)100, j) * pow(1 - p[i] / (double)100, m - j) * comb(m, j);
      g[i][j] = val;

    }
  }

  for(int i = 0; i < n; i++) {
    // i 番目の人が勝つ確率
    double ans = 0.0;

    for(int j = 0; j <= m; j++) {
      // i 番目の人が, j 回休んだ時
      // i 番目の人は it でゴールする
      double mul = 1.0;
      double it = l / v[i] + t[i] * j;

      // 全員に勝たなきゃいけない
      for(int k = 0; k < n; k++) {
        if(i == k) continue;
        double add = 0.0;
        for(int ll = 0;ll <= m; ll++) {
          // k 番目の人は ll 回休む
          // 勝てるなら
          double kt = l / v[k] + t[k] * ll;
          if(it < kt){
            add += g[k][ll];
          }
        }

        mul *= add;
      }

      ans += mul * g[i][j];
    }

    printf("%.10f\n", ans);
  }

  return 0;
}