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 <iostream>
#include <sstream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <ctime>
#include <cstring>
#include <string>
#include <vector>
#include <stack>
#include <queue>
#include <deque>
#include <map>
#include <set>
#include <bitset>
#include <numeric>
#include <utility>
#include <iomanip>
#include <algorithm>
#include <functional>
using namespace std;

typedef long long ll;
typedef vector<int> vint;
typedef vector<long long> vll;
typedef pair<int,int> pint;
typedef pair<long long, long long> pll;

#define MP make_pair
#define PB push_back
#define ALL(s) (s).begin(),(s).end()
#define EACH(i, s) for (__typeof__((s).begin()) i = (s).begin(); i != (s).end(); ++i)
#define COUT(x) cout << #x << " = " << (x) << " (L" << __LINE__ << ")" << endl

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
template<class T1, class T2> ostream& operator << (ostream &s, pair<T1,T2> P) 
{ return s << '<' << P.first << ", " << P.second << '>'; }
template<class T> ostream& operator << (ostream &s, vector<T> P) 
{ for (int i = 0; i < P.size(); ++i) { if (i > 0) { s << " "; } s << P[i]; } return s << endl; }
template<class T1, class T2> ostream& operator << (ostream &s, map<T1,T2> P) 
{ EACH(it, P) { s << "<" << it->first << "->" << it->second << "> "; } return s << endl; }



int N, M, L;
int P[200], T[200], V[200];

double prob[200][200];
double sum[200][200];

void calc_Btry() {
    for (int i = 0; i < N; ++i) {
        prob[i][0] = 1.0;
        for (int j = 1; j <= M; ++j) prob[i][j] = 0.0;
        for (int j = 0; j < M; ++j) {
            for (int k = j; k >= 0; --k) {
                prob[i][k+1] += prob[i][k] * (double)(P[i]) / 100;
                prob[i][k] *= 1.0 - (double)(P[i]) / 100;
            }
        }
        for (int j = 0; j <= M; ++j) sum[i][j] = 0.0;
        sum[i][M] = prob[i][M];
        for (int j = M-1; j >= 0; --j) sum[i][j] = sum[i][j+1] + prob[i][j];
    }
}



int main() {
    while (cin >> N >> M >> L) {
        for (int i = 0; i < N; ++i) cin >> P[i] >> T[i] >> V[i];
        calc_Btry();
        
//        for (int i = 0; i < N; ++i) {
//            cout << i << " : ";
//            for (int j = 0; j <= M; ++j) cout << sum[i][j] << ", " ;
//            cout << endl;
//        }
        
        cout << fixed << setprecision(7);
        for (int i = 0; i < N; ++i) {
            double res = 0.0;
            for (int j = 0; j <= M; ++j) {
                double tmp = prob[i][j];
                double aite = 1.0;
                for (int k = 0; k < N; ++k) {
                    if (k == i) continue;
                    
                    int lim = M;
                    while ( (double)(L)/V[k] + T[k]*lim > (double)(L)/V[i] + T[i]*j ) {
                        if (lim < 0) break;
                        --lim;
                    }
                    ++lim;
                    
                    if (lim > M) aite = 0.0;
                    else aite *= sum[k][lim];
                }
                res += tmp * aite;
            }
            cout << res << endl;
        }
    }
    
    return 0;
}