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 <bits/stdc++.h>
using namespace std;
#define FOR(i,k,n) for(int i = (k); i < (n); i++)
#define REP(i,n) FOR(i,0,n)
#define ALL(a) a.begin(), a.end()
#define MS(m,v) memset(m,v,sizeof(m))
#define D10 fixed<<setprecision(10)
typedef vector<int> vi;
typedef vector<string> vs;
typedef pair<int, int> pii;
typedef long long ll;
typedef long double ld;
const int MOD = 1000000007;
const int INF = MOD + 1;
const ld EPS = 1e-10;
template<class T> T &chmin(T &a, const T &b) { return a = min(a, b); }
template<class T> T &chmax(T &a, const T &b) { return a = max(a, b); }

/*--------------------template--------------------*/
ld dp[55][55];

ld cal(ld p, int n, int k)
{
	if (n == 0) return(k == 0 ? 1.0 : 0);
	if (dp[n][k] >= 0) return dp[n][k];
	ld l = (1 - p)*cal(p, n - 1, k);
	ld r = p*cal(p, n - 1, k - 1);
	return dp[n][k] = l + r;
}


int main()
{
	int n, m;
	ld l;
	cin >> n >> m >> l;
	vector<vector<pair<ld, ld>>> v(n);
	REP(i, n)
	{
		ld p, t, s;
		cin >> p >> t >> s;
		p /= 100;
		if (s == 0)
		{
			v[i].emplace_back(INF, 1.0);
			continue;
		}
		MS(dp, -1);
		REP(j, m + 1) v[i].emplace_back(l / s + t*j, cal(p, m, j));
	}
	vector<ld> ans;
	REP(i, n)
	{
		ld sum = 0;
		REP(j, v[i].size())
		{
			ld ex = v[i][j].second;
			ld t = v[i][j].first;
			REP(k, n)
			{
				if (k == i) continue;
				ld ext = 0;
				REP(kk, v[k].size())
				{
					if (v[k][kk].first > t)
					{
						ext += v[k][kk].second;
					}
				}
				ex *= ext;
			}
			sum += ex;
		}
		ans.push_back(sum);
	}
	REP(i, n) cout << D10 << ans[i] << endl;
	return 0;
}