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;

typedef long long ll;
#define rep(i,n) for(int (i)=0;(i)<(int)(n);++(i))
#define each(itr,c) for(__typeof(c.begin()) itr=c.begin(); itr!=c.end(); ++itr)
#define all(x) (x).begin(),(x).end()
#define pb push_back
#define fi first
#define se second

const double eps=1e-10;

int main()
{
    int n,m,L;
    scanf(" %d %d %d", &n, &m, &L);
    vector<int> p(n),t(n),v(n);
    rep(i,n) scanf(" %d %d %d", &p[i], &t[i], &v[i]);

    // i???????????????j???????????????????¢???????????????¨?????????????????????
    double rest[100][51];
    double total[100][51];

    double dp[51][51];
    rep(i,n)
    {
        memset(dp,0,sizeof(dp));
        dp[0][0]=1;
        rep(j,m)rep(k,j+1)
        {
            dp[j+1][k]+=dp[j][k]*(100-p[i])/100;
            dp[j+1][k+1]+=dp[j][k]*p[i]/100;
        }

        rep(j,m+1)
        {
            rest[i][j]=dp[m][j];
            total[i][j]=(double)L/v[i]+j*t[i];
        }
    }

    double win[100]={0};

    // i???????????????j??????????????????????????????????¢????
    rep(i,n)rep(j,m+1)
    {
        double add=1;
        double mytime=total[i][j];

        rep(k,n)
        {
            if(k==i) continue;

            if(mytime>=total[k][m])
            {
                add=0;
                break;
            }

            int idx=0;
            while(idx<m && mytime>=total[k][idx]) ++idx;

            double q=0;
            for(int l=idx; l<=m; ++l) q+=rest[k][l];

            add*=q;
        }

        win[i]+=add*rest[i][j];
    }

    rep(i,n) printf("%.10f\n", win[i]);
    return 0;
}