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 <cstdio>
#include <vector>
#include <algorithm>
#include <cmath>

#define rep(i,a,n) for(int i=a; i<n; i++)
typedef long long int ll;
const int MAXN = 110;
const ll MAXM = 55;

ll comb[MAXM][MAXM];
double ac[MAXN][MAXM];

int main() {
  rep(i,0,MAXM) comb[i][0] = 1;
  rep(i,1,MAXM) rep(j,1,i+1) comb[i][j] = comb[i-1][j-1] + comb[i-1][j];
  
  int N, M, L;
  scanf("%d %d %d", &N, &M, &L);

  if (N == 0)
    return 1;

  std::vector<int> T(N), V(N);
  std::vector<double> P(N);
  for (int i=0; i<N; ++i) {
    int p;
    scanf("%d %d %d", &p, &T[i], &V[i]);
    P[i] = p / 100.0;
  }

  rep(i,0,N) rep(j,0,M+1) {
    double prev = (j ? ac[i][j-1] : 0);
    ac[i][j] = prev + comb[M][j] * pow(P[i], j) * pow(1-P[i], M-j);
  }

  rep(i,0,N) {
    double ans = 0;
    rep(j,0,M+1) {
      double cur = ac[i][j] - (j ? ac[i][j-1] : 0);
      double user_time = 1.0 * L / V[i] + T[i] * j;
      rep(x,0,N) {
        if(x == i) continue;
        bool flag = false;
        rep(y,0,M+1) {
          double rival_time = 1.0 * L / V[x] + T[x] * y;
          if(user_time < rival_time) {
            flag = true;
            double rival_prob = ac[x][M] - (y ? ac[x][y-1] : 0);
            cur *= rival_prob;
            // printf("user = (%d, %d), rival = (%d, %d), prob = %.12f\n", i+1, j, x+1, y, rival_prob);
            break;
          }
        }
        if(!flag) cur = 0;
      }
      ans += cur;
    }
    printf("%.12f\n", ans);
  }
  return 0;
}