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 fi first
#define se second
#define repl(i,a,b) for(int i=(int)(a);i<(int)(b);i++)
#define rep(i,n) repl(i,0,n)
#define each(itr,v) for(auto itr:v)
#define pb(s) push_back(s)
#define mp(a,b) make_pair(a,b)
#define all(x) (x).begin(),(x).end()
#define dbg(x) cout<<#x"="<<x<<endl
#define maxch(x,y) x=max(x,y)
#define minch(x,y) x=min(x,y)
#define uni(x) x.erase(unique(all(x)),x.end())
#define exist(x,y) (find(all(x),y)!=x.end())
#define bcnt(x) bitset<32>(x).count()

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> P;
typedef pair<P, int> PPI;
typedef pair<ll, ll> PL;
typedef pair<P, ll> PPL;
typedef pair<double,P> DP;

#define INF INT_MAX/3

#define MAX_N 1000

//// combination using double
double comb(int n,int r){
	if(n<r)return 0.0;
	if(n-r<r)r=n-r;
	double res=1;
	rep(i,r){
		res*=n;
		n--;
		res/=i+1;
	}
	return res;
}

int n,m;
double l;
double p[111],t[111],v[111];
DP times[111*55];
double res[111];
double sum[111];

int main(){
	cin.sync_with_stdio(false);
	cin>>n>>m>>l;
	rep(i,n){
		cin>>p[i]>>t[i]>>v[i];
		p[i]/=100;
	}
	int num=n*(m+1);
	rep(i,n)rep(j,m+1){
		times[i*(m+1)+j].fi=(double)j*t[i]+(double)(l/v[i]);
		times[i*(m+1)+j].se=P(i,j);
	}
	sort(times,times+num);
	memset(res,0,sizeof(res));
	rep(i,num){
		double ntimes=times[i].fi;
		int ni=times[i].se.fi,nj=times[i].se.se;
		memset(sum,0,sizeof(sum));
		rep(j,num){
			if(times[j].fi<ntimes||(times[j].se.fi!=ni&&times[j].fi==ntimes))continue;
			int nni=times[j].se.fi,nnj=times[j].se.se;
			if(times[j].se.fi==ni){
				sum[nni]=comb(m,nj)*pow(p[ni],nj)*pow(1-p[ni],m-nj);
			}else{
				sum[nni]+=comb(m,nnj)*pow(p[nni],nnj)*pow(1-p[nni],m-nnj);
			}
		}
		double pp=1;
		rep(i,n)pp*=sum[i];
		res[ni]+=pp;
	}
	rep(i,n)printf("%.10f\n", res[i]);
	return 0;
}