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>
#define ll long long
#define INF 1000000005
#define MOD 1000000007
#define EPS 1e-10
#define rep(i,n) for(int i=0;i<(int)n;++i)
#define all(v) (v).begin(),(v).end()

using namespace std;

typedef pair<int,int>P;

const int MAX_N = 101;
const int MAX_M = 51;

ll comb[MAX_M][MAX_M];
double p[MAX_N];
int t[MAX_N],v[MAX_N];
double bp[MAX_N][MAX_M];
double prt[MAX_N];
double ans[MAX_N];

int main()
{
    int n,m,L;
	cin >> n >> m >> L;
	rep(i,n){
        cin >> p[i] >> t[i] >> v[i];
        p[i] /= 100.0;
	}
    comb[0][0] = 1;
    for(int i = 1;i<=m;i++){
        comb[i][0] = 1;
        for(int j=1;j<=i;j++){
            comb[i][j] = comb[i-1][j-1] + comb[i-1][j];
        }
    }
    rep(i,n){
        if(v[i] == 0){
            prt[i] = INF;
        }else{
            prt[i] = (double)L / v[i];
        }
    }
    rep(i,n){
        rep(j,m+1){
            bp[i][j] = 1.0;
            bp[i][j] *= comb[m][j];
            rep(k,j){
                bp[i][j] *= p[i];
            }
            rep(k,m-j){
                bp[i][j] *= (1.0 - p[i]);
            }
        }
    }
    rep(i,n){
        bool flag = false;
        ans[i] = 0.0;
        rep(j,m+1){
            double tm = bp[i][j];
            rep(k,n){
                double sm = 0.0;
                if(k != i){
                    if(prt[i] + j*t[i] >= prt[k]){
                        if(t[k] == 0){
                            flag = true;
                            break;
                        }
                        int c = (int)((prt[i] + j*t[i] - prt[k] + EPS) / t[k]) + 1;
                        if(c > m){
                            flag = true;
                            break;
                        }
                        for(int l=c;l<=m;l++){
                            sm += bp[k][l];
                        }
                    }else{
                        sm = 1.0;
                    }
                    tm *= sm;
                }
            }
            if(flag){
                break;
            }
            ans[i] += tm;
        }
        printf("%.10lf\n",ans[i]);
    }
	return 0;
}