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 <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>

using namespace std;

#define MAXN 110
#define INF 0x3f3f3f3f
#define PI acos(-1.0)
#define MAX(a,b) (a>b?a:b)
#define MIN(a,b) (a<b?a:b)
#define ABS(m) (m<0?-m:m)
typedef long long LL;

int n,m,l;
LL c[MAXN][MAXN];
int p[MAXN],t[MAXN],v[MAXN];

void Init()
{
    int i,j;
    scanf("%d%d%d",&n,&m,&l);
    for(i=1;i<=n;++i)
        scanf("%d%d%d",p+i,t+i,v+i);
    c[0][0]=1;
    for(i=1;i<=50;++i)
        for(j=1,c[i][0]=1;j<=i;++j)
            c[i][j]=c[i-1][j]+c[i-1][j-1];
}

long double C(long double P,int N,int M)
{
    int i; long double ans=c[N][M];
    for(i=1;i<=M;++i) ans*=P;
    for(;i<=N;++i) ans*=(1-P);
    return ans;
}

void Work()
{
    int i,j,k,x;
    long double ans,r1,r2;
    long double P[MAXN][2];
    LL t1,t2;
    for(i=1;i<=n;++i){
        for(ans=0,j=0;j<=m;++j){
            t1=l+1LL*j*t[i]*v[i];
            r1=C(p[i]/100.0,m,j);
            for(k=1;k<=n;++k){
                if(k==i) continue;
                for(x=1,P[0][0]=P[0][1]=1;x<=m;++x)
                    P[x][0]=P[x-1][0]*p[k]/100.0,P[x][1]=P[x-1][1]*(1-p[k]/100.0);
                for(x=r2=0,t2=l;x<=m&&t2*v[i]<=t1*v[k];t2+=1LL*t[k]*v[k],++x);
                for(r2=0;x<=m;++x) r2+=((long double)c[m][x])*P[x][0]*P[m-x][1];
                r1*=r2;
            }
            ans+=r1;
        }
        printf("%.12f\n",(double)ans);
    }
}

int main()
{
    Init();
    Work();
    return 0;
}