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:
#define __USE_MINGW_ANSI_STDIO 0
#include <bits/stdc++.h>

using namespace std;
using ll = long long;
#define int ll
using VI = vector<int>;
using VVI = vector<VI>;
using PII = pair<int, int>;

#define FOR(i, a, n) for (ll i = (ll)a; i < (ll)n; ++i)
#define REP(i, n) FOR(i, 0, n)
#define ALL(x) x.begin(), x.end()
#define PB push_back

const ll LLINF = (1LL<<60);
const int INF = (1LL<<30);
const int MOD = 1000000007;

template <typename T> T &chmin(T &a, const T &b) { return a = min(a, b); }
template <typename T> T &chmax(T &a, const T &b) { return a = max(a, b); }
template <typename T> bool IN(T a, T b, T x) { return a<=x&&x<b; }
template<typename T> T ceil(T a, T b) { return a/b + !!(a%b); }
template<class S,class T>
ostream &operator <<(ostream& out,const pair<S,T>& a){
  out<<'('<<a.first<<','<<a.second<<')';
  return out;
}
template<class T>
ostream &operator <<(ostream& out,const vector<T>& a){
  out<<'[';
  REP(i, a.size()) {out<<a[i];if(i!=a.size()-1)out<<',';}
  out<<']';
  return out;
}

int dx[] = {0, 1, 0, -1}, dy[] = {1, 0, -1, 0};

ll combi(ll n, ll k) {
  static ll c[55][55];
  if(c[0][0] == 0) {
    REP(i, 55) {
      c[i][0] = c[i][i] = 1;
      FOR(j, 1, i) c[i][j] = c[i-1][j] + c[i-1][j-1];
    }
  }
  return c[n][k];
}

pair<double, double> dp[105][55];
int p[105], t[105], v[105];
signed main(void)
{
  cin.tie(0);
  ios::sync_with_stdio(false);

  int n, m, L;
  cin >> n >> m >> L;
  REP(i, n) cin >> p[i] >> t[i] >> v[i];

  combi(2, 0);

  REP(i, n) {
    REP(j, m+1) {
      dp[i][j] = {j*t[i] + (double)L/v[i], combi(m, j) * pow(p[i]/100.0, j) * pow((100-p[i])/100.0, m-j)};
      // cout << (double)combi(m, j) << "," << pow(p[i]/100.0, j) << "," << pow((100-p[i])/100.0, m-j) << "," << (100-p[i])/100.0 << " ";
      // cout << dp[i][j] << " ";
    }
    // cout << endl;
  }

  vector<double> ans(n, 0);
  // i人目がj回休んだ時に優勝する確率
  REP(i, n) {
    REP(j, m+1) {
      double prob = dp[i][j].second;
      REP(k, n) {
        if(i==k) continue;
        double sum = 0;
        for(int l=m; l>=0; --l) {
          if(dp[k][l].first <= dp[i][j].first) {
            prob *= sum;
            break;
          }
          sum += dp[k][l].second;
        }
      }
      ans[i] += prob;
    }
  }

  cout << fixed << setprecision(9);
  REP(i, n) cout << ans[i] << endl;

  return 0;
}