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 int long long
typedef vector<int>vint;
typedef pair<int,int>pint;
typedef vector<pint>vpint;
#define rep(i,n) for(int i=0;i<(n);i++)
#define reps(i,f,n) for(int i=(f);i<(n);i++)
#define all(v) (v).begin(),(v).end()
#define each(it,v) for(__typeof((v).begin()) it=(v).begin();it!=(v).end();it++)
#define pb push_back
#define fi first
#define se second
template<typename A,typename B>inline void chmin(A &a,B b){if(a>b)a=b;}
template<typename A,typename B>inline void chmax(A &a,B b){if(a<b)a=b;}

int N,M,L;
double P[100],T[100],V[100];

double p[100][55],t[100][55];

double dp[55][55];

signed main(){
    cin>>N>>M>>L;
    rep(i,N)cin>>P[i]>>T[i]>>V[i],P[i]/=100;

    rep(i,N){
        memset(dp,0,sizeof(dp));
        dp[0][0]=1.0;
        for(int j=0;j<M;j++){
            for(int k=0;k<=j;k++){
                dp[j+1][k+1]+=dp[j][k]*P[i];
                dp[j+1][k]+=dp[j][k]*(1-P[i]);
            }
        }

        for(int j=0;j<=M;j++){
            t[i][j]=L/V[i]+j*T[i];
            p[i][j]=dp[M][j];
        }
    }

    rep(i,N){
        double latte=0.0;
        for(int j=0;j<=M;j++){
            double malta=p[i][j];
            for(int k=0;k<N;k++)if(k!=i){
                double mocha=0.0;
                for(int l=0;l<=M;l++)if(t[k][l]>t[i][j])mocha+=p[k][l];
                malta*=mocha;
            }
            latte+=malta;
        }
        printf("%.20f\n",latte);
    }
    return 0;
}