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"

using namespace std;

using ll = long long;
using ld = long double;
using P = pair<int, int>;
constexpr ld EPS = 1e-12;
constexpr int INF = numeric_limits<int>::max() / 2;
constexpr int MOD = 1e9 + 7;

int main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);

    // N 人, M 箇所, 距離 L
    int N, M, L;
    cin >> N >> M >> L;
    // 休憩確率, 休憩時間, 速度
    vector<ld> P(N), T(N), V(N);
    for (int i = 0; i < N; i++)
    {
        cin >> P[i] >> T[i] >> V[i];
        P[i] /= 100;
    }
    // comb[i][j] := iCj
    ld comb[60][60];
    for (int i = 0; i <= 50; i++)
        comb[i][0] = 1.0;
    for (int i = 1; i <= 50; i++)
    {
        for (int j = 1; j <= i; j++)
        {
            comb[i][j] = comb[i - 1][j - 1] + comb[i - 1][j];
        }
    }
    // pn[i][j] := i 人目が j 回休憩する確率
    // tn[i][j] := i 人目が j 回休んだ時の所要時間 (L/V[i] + j*T)
    vector<vector<ld>> pn(N, vector<ld>(M + 1)), tn(N, vector<ld>(M + 1));
    for (int i = 0; i < N; i++)
    {
        for (int j = 0; j <= M; j++)
        {
            tn[i][j] = ld(L) / V[i] + j * T[i];
            pn[i][j] = comb[M][j] * pow(P[i], j) * pow(1.0 - P[i], M - j);
        }
    }
    for (int i = 0; i < N; i++)
    {
        if (abs(V[i]) < EPS)
        {
            cout << fixed << setprecision(10) << 0.0 << endl;
            continue;
        }
        ld res = 0;
        for (int j = 0; j <= M; j++)
        {
            ld p = pn[i][j];
            for (int k = 0; k < N; k++)
            {
                if (k == i)
                    continue;
                if (abs(V[i]) < EPS)
                    continue;
                ld sum = 0;
                for (int l = 0; l <= M; l++)
                {
                    if (tn[k][l] > tn[i][j] + EPS)
                        sum += pn[k][l];
                }
                p *= sum;
            }
            res += p;
        }
        cout << fixed << setprecision(10) << res << endl;
    }
}