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 pii = pair<int, int>;
using pll = pair<ll, ll>;

#define FOR(k,m,n) for(ll (k)=(m);(k)<(n);(k)++)
#define REP(i,n) FOR((i),0,(n))
#define WAITING(str) int str;std::cin>>str;
#define DEBUGING(str) cout<< #str << " " str<<endl

constexpr int INF = (1 << 30);
constexpr ll INFL = (1ll << 60);
constexpr ll MOD = 1000000007;// 10^9+7

using ar3 = array<int, 3>;

ld comb(ll l, ll r) {
	ld res = 1;
	REP(i, r) {
		res *= l - i;
		res /= i + 1;
	}
	return res;
}

// [person][rest-number] -> prob
vector<vector<ld>> calc_prob_table(const vector<ar3>& info, const int M) {
	const int N = info.size();
	vector<vector<ld>> res(N, vector<ld>(M + 1));
	REP(i, N)REP(j, M + 1) {
		ld rest = info[i][0];
		res[i][j] = comb(M, j)*powl(rest / 100., j)*powl(1 - rest / 100., M - j);
	}
	//REP(i, N)REP(j, M - 1)res[i][j + 1] += res[i][j];
	return res;
}

ld lose_prob(const vector<ld>& prob, ar3 info, ld time, const ld L) {
	const int M = prob.size();
	time -= L / info[2];
	if (info[1] == 0) {
		time = INF * (time >= 0 ? 1 : -1);
	}
	else {
		time /= (ld)info[1];
	}

	ld res = 0;
	REP(i, min(time + 1e-5, (ld)M)) {
		res += prob[i];
	}
	return 1 - res;
}

int main()
{
	int N, M, L;
	cin >> N >> M >> L;
	vector<ar3> info(N);
	REP(i, N)REP(j, 3)cin >> info[i][j];

	if (N == 1) {
		cout << 1 << endl;
		return 0;
	}

	auto table = calc_prob_table(info, M);
	REP(i, N) {
		ld res = 0;
		REP(rest, M + 1) {
			if (info[i][2] == 0)break;
			ld tres = table[i][rest];
			ld time = (ld)L / info[i][2] + info[i][1] * rest;
			REP(opp, N)if (opp != i && info[opp][2] != 0) {
				tres *= lose_prob(table[opp], info[opp], time, L);
			}
			res += tres;
		}
		cout << fixed << setprecision(10) << res << endl;
	}

	return 0;
}