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 <cstdio>
#include <cmath>
#define REP(i, a, n) for(int i = ((int) a); i < ((int) n); i++)
#define INF 1e15
using namespace std;
typedef long long ll;

int N, M, L, P[100], T[100], V[100];
double p[100][51];

ll combination(ll n, ll r) {
  static ll comb[51][51];
  ll &ret = comb[n][r];
  if(ret != 0) return ret;
  if(n == 0 || r == 0) return ret = 1;
  if(n == r) return ret = 1;
  if(r == 1) return ret = n;
  return ret = combination(n - 1, r - 1) + combination(n - 1, r);
}

bool ok(int ai, int aj, int bi, int bj) {
  double ta = V[ai] > 0 ? (double) L / V[ai] + T[ai] * aj : INF;
  double tb = V[bi] > 0 ? (double) L / V[bi] + T[bi] * bj : INF;
  return ta < tb;
}

int main(void) {
  cin >> N >> M >> L;
  REP(i, 0, N) cin >> P[i] >> T[i] >> V[i];

  REP(i, 0, N) REP(j, 0, M + 1) {
    p[i][j] = combination(M, j) * pow(P[i] / 100.0, j) * pow(1 - P[i] / 100.0, M - j);
  }

  REP(i, 0, N) {
    double ans = 0;
    double pi = 0;
    REP(j, 0, M + 1) {
      pi = p[i][j];
      double pij = pi;
      REP(k, 0, N) if(i != k) {
        double pk = 0;
        REP(l, 0, M + 1) if(ok(i, j, k, l)) pk += p[k][l];
        pij *= pk;
      }
      ans += pij;
    }
    printf("%.8lf\n", ans);
  }


  return 0;
}