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:
// #define _GLIBCXX_DEBUG // for STL debug (optional)
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <string>
#include <cstring>
#include <deque>
#include <list>
#include <queue>
#include <stack>
#include <vector>
#include <utility>
#include <algorithm>
#include <map>
#include <set>
#include <complex>
#include <cmath>
#include <limits>
#include <cfloat>
#include <climits>
#include <ctime>
#include <cassert>
#include <numeric>
#include <fstream>
#include <functional>
#include <bitset>
using namespace std;
using ll = long long int;
using int64 = long long int;
 
template<typename T> void chmax(T &a, T b) {a = max(a, b);}
template<typename T> void chmin(T &a, T b) {a = min(a, b);}
template<typename T> void chadd(T &a, T b) {a = a + b;}
 
int dx[] = {0, 0, 1, -1};
int dy[] = {1, -1, 0, 0};
const int INF = 1001001001;
const ll LONGINF = 1001001001001001LL;
const ll MOD = 1000000007LL;

const double EPS = 1e-12;
double dp[110][51][51];
int main() {
    int N, M, L; cin >> N >> M >> L;
    vector<int> P(N), T(N), V(N);
    for(int i=0; i<N; i++) {
        cin >> P[i] >> T[i] >> V[i];
    }

    for(int x=0; x<N; x++) {
        dp[x][0][0] = 1.0;
        double p = 1.0 * P[x] / 100;
        
        for(int i=0; i<=M; i++) {
            for(int j=0; j<=M; j++) {
                // fprintf(stderr, "dp[%d][%d][%d] = %.12f\n", x, i, j, dp[x][i][j]);
                
                if(i == M or dp[x][i][j] < EPS) continue;
                // やすむ
                dp[x][i+1][j+1] += dp[x][i][j] * p;

                // やすまない
                dp[x][i+1][j  ] += dp[x][i][j] * (1 - p);
            }
        }

        // るいせき
        for(int j=1; j<=M; j++) dp[x][M][j] += dp[x][M][j-1];
    }

    for(int x=0; x<N; x++) {
        // x 番目の人が i 回休んだ場合
        double ans = 0.0;
        for(int i=0; i<=M; i++) {
            double p = (i ? dp[x][M][i] - dp[x][M][i-1] : dp[x][M][i]);
            double tx = 1.0 * L / V[x] + T[x] * i;
            
            // y 番目の人が j 回休む
            for(int y=0; y<N; y++) {
                if(x == y) continue;

                bool defeat = false;
                for(int j=0; j<=M; j++) {
                    double ty = 1.0 * L / V[y] + T[y] * j;
                    if(ty - tx > EPS) {
                        // fprintf(stderr, "x = %d, i = %d, y = %d, j = %d: tx = %.12f, ty = %.12f\n", x, i, y, j, tx, ty);
                        // j 回以上休んだら勝てる
                        double q = (j ? dp[y][M][j-1] : 0);
                        p *= (1 - q);
                        defeat = true;
                        break;
                    }
                }

                if(!defeat) p = 0;
            }
            ans += p;
        }
        printf("%.12f\n", ans);
    }
    return 0;
}