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 <vector>
#include <list>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <deque>
#include <algorithm>
#include <utility>
#include <functional>
#include <sstream>
#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <climits>
#include <cassert>

using namespace std;

typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<string> vs;
typedef pair<int,int> pii;
typedef long long ll;

#define ALL(a) (a).begin(),(a).end()
#define RALL(a) (a).rbegin(),(a).rend()
#define EXIST(s,e) ((s).find(e)!=(s).end())
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n) FOR(i,0,n)

#define PB push_back

typedef vector<double> vd;

const ll M=1000000000;

class Runner {
public:
	vd goal_times;
	vd probabilities;

	Runner() {
		goal_times = vd();
		probabilities = vd();
	}

	void inspect() {
		cerr << "goal_times" << endl;
		REP(i, goal_times.size()) {
			cerr << " " << goal_times[i];
		}
		cerr << endl;
		cerr << "probabilities" << endl;
		REP(i, probabilities.size()) {
			cerr << " " << probabilities[i];
		}
		cerr << endl;
	}
};


ll combination(int n,int k){
	ll ans=1;
	FOR(i,1,k+1){
		ans*=n-i+1;
		ans/=i;
	}
	return ans;
}

const double EPS = 1e-9;
int main()
{
	cout.precision(16);
	int N, M, L; cin >> N >> M >> L;
	vector<Runner> runners;
	REP(i, N) {
		Runner runner;
		double P; int T, V; cin >> P >> T >> V; P /= 100.0;
		double base_time = V > 0 ? (double)L / V : 1e20;
		REP(t_break, M + 1) {
			runner.goal_times.PB(base_time + T * t_break);
			runner.probabilities.PB(
				pow(P, t_break) * pow(1 - P, M - t_break) * combination(M, t_break) );
		}
		runners.PB(runner);
		//runner.inspect();
	}

	REP(i, runners.size()) {
		Runner &r = runners[i];
		double ans = 0;
		REP(t, M + 1) {
			double prize_prob = r.probabilities[t];
			REP(j, runners.size()) {
				if (i == j) continue;
				Runner &comp = runners[j];
				double win_prob = 0;
				REP(k, M+1) {
					if(comp.goal_times[k] > r.goal_times[t] + EPS) {
						win_prob += comp.probabilities[k];
					}
				}
				prize_prob *= win_prob;
			}
			ans += prize_prob;
		}
		printf("%.8f\n",ans);
		//cout << ans << endl;
	}
}