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<string>
#include<cstdio>
#include<vector>
#include<cmath>
#include<algorithm>
#include<functional>
#include<iomanip>
#include<queue>
#include<ciso646>
#include<random>
#include<map>
#include<set>
#include<complex>
#include<bitset>
#include<stack>
#include<unordered_map>
using namespace std;
typedef long long ll;
typedef unsigned int ui;
const ll mod = 1000000007;
const ll INF = (ll)1000000007 * 1000000007;
typedef pair<int, int> P;
#define stop char nyaa;cin>>nyaa;
#define rep(i,n) for(int i=0;i<n;i++)
#define per(i,n) for(int i=n-1;i>=0;i--)
#define Rep(i,sta,n) for(int i=sta;i<n;i++)
#define rep1(i,n) for(int i=1;i<=n;i++)
#define per1(i,n) for(int i=n;i>=1;i--)
#define Rep1(i,sta,n) for(int i=sta;i<=n;i++)
typedef long double ld;
typedef complex<ld> Point;
const ld eps = 1e-8;
const ld pi = acos(-1.0);
typedef pair<ll, ll> LP;
typedef pair<ld, ld> LDP;
struct runner { ld p, t, v; };
runner a[100];
ld pas[100][100];
void init() {
	rep(i, 100) {
		rep(j, 100) {
			if (i == 0 || j == 0)pas[i][j] = 1;
			else pas[i][j] = pas[i - 1][j] + pas[i][j - 1];
		}
	}
}
//i番目の人がj回休憩をとる確率
ld pr[100][51];
int main(){
	init();
	cout << fixed << setprecision(10);
	int n, m;ld L; cin >> n >> m >> L;
	rep(i, n) {
		ld p, t, v; cin >> p >> t >> v;
		p /= 100.0;
		a[i] = { p,t,v };
		rep(j, m + 1) {
			pr[i][j] = pas[m - j][j];
			rep(k, j) {
				pr[i][j] *= p;
			}
			rep(k, m-j) {
				pr[i][j] *= (1 - p);
			}
		}
	}
	rep(i, n) {
		ld out = 0;
		rep(j, m + 1) {
			ld sum = j * a[i].t + L / a[i].v;
			ld ssum = pr[i][j];
			rep(k, n) {
				if (i == k)continue;
				ld csum = 0;
				rep(l, m+1) {
					ld pp = l * a[k].t + L / a[k].v;
					if (pp > sum) {
						csum += pr[k][l];
					}
				}
				ssum *= csum;
			}
			out += ssum;
		}
		cout << out << endl;
	}
	return 0;
}