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 <algorithm>
#include <iomanip>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <numeric>
#include <bitset>
#include <cmath>

static const int MOD = 1000000007;
using ll = long long;
using u32 = uint32_t;
using namespace std;

template<class T> constexpr T INF = ::numeric_limits<T>::max() / 32 * 15 + 208;

int main() {
    int n, m, l;
    cin >> n >> m >> l;
    using P = pair<double, double>;
    vector<vector<P>> v(n, vector<P>(m+1, {0,0}));
    for (int i = 0; i < n; ++i) {
        double p, t, k;
        cin >> p >> t >> k;
        p /= 100; k = l/k;
        vector<vector<double>> dp(m+1, vector<double>(m+1, 0));
        dp[0][0] = 1.0;
        for (int j = 0; j < m; ++j) {
            for (int jj = 0; jj <= j; ++jj) {
                dp[j+1][jj] += dp[j][jj]*(1-p);
                dp[j+1][jj+1] += dp[j][jj]*p;
            }
        }
        for (int j = 0; j <= m; ++j) {
            v[i][m-j] = make_pair(-k-t*j, dp[m][j]);
        }
        for (int j = 0; j <= m; ++j) {
            if(j) v[i][j].second += v[i][j-1].second;
        }
    }
    for (int i = 0; i < n; ++i) {
        double ans = 0;
        for (int j = 0; j <= m; ++j) {
            double x = v[i][j].second;
            if(j) x -= v[i][j-1].second;
            for (int k = 0; k < n; ++k) {
                if(i == k) continue;
                auto cur = lower_bound(v[k].begin(),v[k].end(), make_pair(v[i][j].first, 0.0));
                if(cur == v[k].begin()) x *= 0;
                else {
                    cur--;
                    x *= cur->second;
                }
            }
            ans += x;
        }
        printf("%.10lf\n", ans);
    }
    return 0;
}