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;

inline int toInt(string s) {int v; istringstream sin(s);sin>>v;return v;}
template<class T> inline string toString(T x) {ostringstream sout;sout<<x;return sout.str();}
template<class T> inline T sqr(T x) {return x*x;}

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 pb push_back
#define mp make_pair
#define each(i,c) for(typeof((c).begin()) i=(c).begin(); i!=(c).end(); ++i)
#define exist(s,e) ((s).find(e)!=(s).end())
#define range(i,a,b) for(int i=(a);i<(b);++i)
#define rep(i,n)  range(i,0,n)
#define clr(a,b) memset((a), (b) ,sizeof(a))
#define dump(x)  cerr << #x << " = " << (x) << endl;
#define debug(x) cerr << #x << " = " << (x) << " (L" << __LINE__ << ")" << " " << __FILE__ << endl;

const double eps = 1e-10;
const double pi  = acos(-1.0);
const ll INF =1LL << 62;
const int inf =1 << 29;

int n,m,l;
int p[110],t[110],v[110];

double rt[110][55];
double rp[110][55];

double com[55][55];
double ans[110];

void init(){
	rep(i,55)rep(j,i+1){
		if(j==0||j==i)
			com[i][j]=1.0;
		else
			com[i][j]=com[i-1][j]+com[i-1][j-1];
	}
}

int main(void){
	init();
	cin >> n >> m >> l;
	rep(i,n) cin >> p[i] >> t[i] >> v[i];
	rep(i,n)rep(j,m+1){
		rt[i][j]=1.0*l/v[i]+1.0*j*t[i];
		rp[i][j]=com[m][j]*pow(1.0*p[i]/100.0,j)*pow( (100.0-p[i])/100.0,m-j);
	}

	rep(i,n){
		ans[i]=0.0;
		rep(j,m+1){
			double prod=1.0;
			rep(a,n){
				if(a==i) continue;
				double sum=0.0;
				rep(b,m+1) if(rt[i][j]<rt[a][b]) sum+=rp[a][b];
				prod*=sum;
			}
			ans[i]+=rp[i][j]*prod;
		}
	}


	cout.precision(20);
	rep(i,n) cout << fixed << ans[i]  << endl;
	return 0;
}