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 <cstdlib>
#include <cstring>
#include <cmath>
#include <climits>
#include <cfloat>
#include <ctime>
#include <cassert>
#include <map>
#include <utility>
#include <set>
#include <iostream>
#include <memory>
#include <string>
#include <vector>
#include <algorithm>
#include <functional>
#include <sstream>
#include <complex>
#include <stack>
#include <queue>
#include <numeric>
#include <list>
#include <iomanip>
#include <fstream>
#include <bitset>
   
using namespace std;
   
 
#define rep(i, n) for (int i = 0; i < (int)(n); ++i)
#define foreach(it, c) for (__typeof__((c).begin()) it=(c).begin(); it != (c).end(); ++it)
#define rforeach(it, c) for (__typeof__((c).rbegin()) it=(c).rbegin(); it != (c).rend(); ++it)
#define all(c) (c).begin(), (c).end()
#define rall(c) (c).rbegin(), (c).rend()
#define CL(arr, val) memset(arr, val, sizeof(arr))
#define COPY(dest, src) memcpy(dest, src, sizeof(dest))
#define ten(n) ((long long)(1e##n))
#define bin(n) (1LL << (n))
#define erep(i, n) for (int i = 0; i <= (int)(n); ++i)
#define revrep(i, n) for (int i = (n); i >= 0; --i)
#define pb push_back

template <class T> void chmax(T& a, const T& b) { a = max(a, b); }
template <class T> void chmin(T& a, const T& b) { a = min(a, b); }

template <class T> void uniq(T& c) { sort(c.begin(), c.end()); c.erase(unique(c.begin(), c.end()), c.end()); }
 
template <class T> string to_s(const T& a) { ostringstream os; os << a; return os.str(); }
template <class T> T to_T(const string& s) { istringstream is(s); T res; is >> res; return res; }

template <typename T> void print_container(ostream& os, const T& c) { const char* _s = " "; if (!c.empty()) { __typeof__(c.begin()) last = --c.end(); foreach (it, c) { os << *it; if (it != last) cout << _s; } } }
template <typename T> ostream& operator<<(ostream& os, const vector<T>& c) { print_container(os, c); return os; }
template <typename T> ostream& operator<<(ostream& os, const set<T>& c) { print_container(os, c); return os; }
template <typename T> ostream& operator<<(ostream& os, const multiset<T>& c) { print_container(os, c); return os; }
template <typename T> ostream& operator<<(ostream& os, const deque<T>& c) { print_container(os, c); return os; }
template <typename T, typename U> ostream& operator<<(ostream& os, const map<T, U>& c) { print_container(os, c); return os; }
template <typename T, typename U> ostream& operator<<(ostream& os, const pair<T, U>& p) { os << "( " << p.first << ", " << p.second << " )"; return os; }

template <class T> void print(T a, int n, const string& deli = " ", int br = 1) { for (int i = 0; i < n; ++i) { cout << a[i]; if (i + 1 != n) cout << deli; } while (br--) cout << endl; }
template <class T> void print2d(T a, int w, int h, int width = -1, int br = 1) { for (int i = 0; i < h; ++i) { for (int j = 0; j < w; ++j) {    if (width != -1) cout.width(width); cout << a[i][j] << ' '; } cout << endl; } while (br--) cout << endl; }

template <class T> void input(T& a, int n) { for (int i = 0; i < n; ++i) cin >> a[i]; }
template <class T> void input(T* a, int n) { for (int i = 0; i < n; ++i) cin >> a[i]; }

void fix_pre(int n) { cout.setf(ios::fixed, ios::floatfield); cout.precision(10); }
void fast_io() { cin.tie(0); ios::sync_with_stdio(false); }
#define trace(x) (cout << #x << ": " << (x) << endl)
 
bool in_rect(int x, int y, int w, int h) { return 0 <= x && x < w && 0 <= y && y < h; }
bool in_seg(int n, int l, int r) { return l <= n && n < r; } // n in [l, r)?

typedef long long ll;
typedef pair<int, int> pint;
 
// y(v): v>^<  y(^): ^>v<
const int dx[] = { 0, 1, 0, -1 };
const int dy[] = { 1, 0, -1, 0 };

const double PI = acos(-1.0);
#define mp make_pair


int n, m, l;
int t[111], v[111];
double p[111];
bool comp(pint a, pint b)
{
    int i = a.first, j = b.first;
    int ra = a.second, rb = b.second;

    int x = v[j] * (v[i] * ra * t[i] + l);
    int y = v[i] * (v[j] * rb * t[j] + l);
    return x > y;
}
int main()
{
    cin >> n >> m >> l;
    rep(i, n)
        cin >> p[i] >> t[i] >> v[i], p[i] /= 100;

    vector<pint> time; // (index, rest)
    rep(i, n) erep(j, m)
        time.pb(pint(i, j));
    sort(all(time), comp);
    int rank[111][55]; // (index, rest) -> rank
    rep(i, time.size())
        rank[time[i].first][time[i].second] = i;
    
    double c[111][111];
    CL(c, 0);
    erep(i, m)
    {
        c[i][0] = 1;
        for (int j = 1; j <= i; ++j)
            c[i][j] = c[i - 1][j] + c[i - 1][j - 1];
    }

    double run_p[111][55]; // (index, rest) -> prob
    double sum[111][111 * 55]; // (index, rank) -> [0, rank)sum prob
    CL(sum, 0);
    rep(i, n)
    {
        double pp[55], np[55];
        pp[0] = np[0] = 1;
        rep(j, m)
        {
            pp[j + 1] = pp[j] * p[i];
            np[j + 1] = np[j] * (1 - p[i]);
        }

        erep(j, m)
            run_p[i][j] = c[m][j] * pp[j] * np[m - j];

        erep(j, m)
            sum[i][rank[i][j] + 1] = run_p[i][j];
        rep(j, time.size())
            sum[i][j + 1] += sum[i][j];
    }

    rep(i, n)
    {
        double res = 0;
        erep(j, m)
        {
            int k = lower_bound(all(time), pint(i, j), comp) - time.begin();

            double win_p = 1;
            rep(w, n)
                if (w != i)
                    win_p *= sum[w][k];

            res += run_p[i][j] * win_p;
        }
        printf("%.10f\n", res);
    }
}