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 <vector>
#include <algorithm>
#include <cstdio>

using namespace std;

struct Tag {
    double p, t, v;
};

double dp[51][100][51]; // dp[M][N][k] for k rests
double tim[100][51];

int main() {
    int N, M, L;
    cin >> N >> M >> L;

    vector<Tag> v;
    for(int i = 0; i < N; ++i) {
        Tag t;
        cin >> t.p >> t.t >> t.v;
        t.p /= 100;
        v.push_back(t);
    }

    for(int i = 0; i < N; ++i) {
        dp[0][i][0] = 1;
    }
    for(int m = 1; m <= M; ++m) {
        for(int n = 0; n < N; ++n) {
            for(int k = 0; k <= M; ++k) {
                dp[m][n][k] = v[n].p*dp[m-1][n][k-1] + (1-v[n].p)*dp[m-1][n][k];
                //printf("%d %d %d = %d\n", m, n, k, dp[m][n][k]);
            }
        }
    }
    for(int n = 0; n < N; ++n) {
        tim[n][0] = (double)L / v[n].v;
        for(int k = 1; k <= M; ++k) {
            tim[n][k] = tim[n][k-1] + v[n].t;
        }
    }
    for(int n = 0; n < N; ++n) {
        double prob = 0;
        if(v[n].v == 0) goto END;

        for(int k = 0; k <= M; ++k) {
            double tmp = dp[M][n][k];
            for(int nn = 0; nn < N; ++nn) {
                if(n == nn) continue;
                double *bound = upper_bound(tim[nn], tim[nn]+M+1, tim[n][k]);
                int idx = bound - tim[nn];
                double sum = 0;
                for(int i = idx; i <= M; ++i) sum += dp[M][nn][i];
                tmp *= sum;
            }
            prob += tmp;
        }
END:
        printf("%.6f\n", prob);
    }

    return 0;
}