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 <algorithm>
#include <cmath>
#include <climits>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <fstream>
#include <iostream>
#include <list>
#include <map>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <vector>
#include <cassert>
#include <functional>

using namespace std;

#define LOG(...) printf(__VA_ARGS__)
//#define LOG(...)
#define FOR(i,a,b) for(int i=(int)(a);i<(int)(b);++i)
#define REP(i,n) for(int i=0;i<(int)(n);++i)
#define ALL(a) (a).begin(),(a).end()
#define RALL(a) (a).rbegin(),(a).rend()
#define EXIST(s,e) ((s).find(e)!=(s).end())
#define SORT(c) sort((c).begin(),(c).end())
#define RSORT(c) sort((c).rbegin(),(c).rend())
#define CLR(a) memset((a), 0 ,sizeof(a))

typedef long long ll;
typedef unsigned long long ull;
typedef vector<bool> vb;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector<vb> vvb;
typedef vector<vi> vvi;
typedef vector<vll> vvll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;

const int dx[] = { -1, 0, 1, 0 }; const int dy[] = { 0, 1, 0, -1 };

struct UnionFind {
	vector<int> v;
	UnionFind(int n) : v(n)  { for (int i = 0; i < n; i++) v[i] = i; }
	int  find(int x)         { return v[x] == x ? x : v[x] = find(v[x]); }
	void unite(int x, int y) { v[find(x)] = find(y); }
};
ll conb(int a, int b) {
	ll p = 1;
	FOR(i, 1, b+1) {
		p = p*(a - i + 1) / i;
	}
	return p;
}


int main() {
	int n, m;
	double l;
	cin >> n >> m >> l;
	vector<vector<pair<double, double>>> goal(n,vector<pair<double,double>>(m+1));//time ,prob
	REP(i, n) {
		double pro,time,v;
		cin>> pro>> time>> v;
		pro /= 100.0;
		REP(j,m+1) {
			goal[i][j] = {l/v+time*(j),(double)conb(m,j)*pow(pro,j)*pow(1.0-pro,m-j)};
		}
	}
	REP(i, n) {
		double ans=0.0;
		REP(j, m+1) {
			double res=goal[i][j].second;
			bool win=true;
			REP(k, n) {
				if (i == k)
					continue;
				double res2=1.0;
				REP(l,m+1) {
					if (goal[i][j].first<goal[k][l].first-1e-9) {
						res *= res2;
						break;
					}
					else {
						res2 -= goal[k][l].second;
						if (l == m)
							win = false;
					}
				}
			}
			if(win)
			ans += res;
		}
		printf("%.8lf\n", ans);
	}
}