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 <cmath>
#include <algorithm>
using namespace std;

const int MAX_N = 100;
const int MAX_M = 50;
long long int comb[MAX_M+5][MAX_M+5];
int P[MAX_N+5], T[MAX_N+5], V[MAX_N+5];
double prob[MAX_N+5][MAX_M+5];
double goal_t[MAX_N+5][MAX_M+5];

void init_comb() {
    comb[0][0] = 1;
    for(int i=1; i<=MAX_M; i++) {
        comb[i][0] = 1;
        for(int j=1; j<=MAX_M; j++) {
            comb[i][j] = comb[i-1][j-1] + comb[i-1][j];
        }
    }
}

int main() {
    init_comb();
    int N, M, L; scanf("%d%d%d", &N, &M, &L);
    for(int i=0; i<N; i++) {
        scanf("%d%d%d", &P[i], &T[i], &V[i]);
        for(int k=0; k<=M; k++) {
            goal_t[i][k] = 1.0 * L / V[i] + k * T[i];
            prob[i][k] = comb[M][k] * pow(P[i] / 100.0, k) * pow((100-P[i]) / 100.0, M-k);
            if(k != 0) prob[i][k] += prob[i][k-1];
        }
    }

    for(int i=0; i<N; i++) {
        double ans = 0.0;
        for(int k=0; k<=M; k++) {
            double player_t = goal_t[i][k], tmp = prob[i][k] - (k != 0 ? prob[i][k-1] : 0);
            for(int j=0; j<N; j++) {
                if(i == j) continue;
                // かてない
                if(player_t > goal_t[j][M]) tmp = 0;
                // よゆう
                else if(player_t < goal_t[j][0]) continue;
                else {
                    int idx = upper_bound(goal_t[j], goal_t[j]+M+1, player_t) - goal_t[j] - 1;
                    tmp *= (1.0 - prob[j][idx]);
                }
            }
            // printf("initial tmp = %.12f, current = %.12f\n", prob[i][k], tmp);
            ans += tmp;
        }
        printf("%.12f\n", ans);
    }
    return 0;
}