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:
#define  _CRT_SECURE_NO_WARNINGS
#pragma comment (linker, "/STACK:526000000")

#include "bits/stdc++.h"

using namespace std;
typedef string::const_iterator State;
#define eps 1e-11L
#define MAX_MOD 1000000007LL
#define GYAKU 500000004LL

#define MOD 998244353LL
#define seg_size 262144 * 4LL
#define pb push_back
#define mp make_pair
typedef long long ll;
#define REP(a,b) for(long long (a) = 0;(a) < (b);++(a))
#define ALL(x) (x).begin(),(x).end()

unsigned long xor128() {
	static unsigned long x = 123456789, y = 362436069, z = 521288629, w = time(NULL);
	unsigned long t = (x ^ (x << 11));
	x = y; y = z; z = w;
	return (w = (w ^ (w >> 19)) ^ (t ^ (t >> 8)));
}
void init() {
	iostream::sync_with_stdio(false);
	cout << fixed << setprecision(20);
}

#define int ll

long double comb(int a, int b) {
	long double ans = 1.0L;
	for (int i = 0; i < b; ++i) {
		ans *= (long double)(a - i);
		ans /= (long double)(i + 1);
	}
	return ans;
}

long double prob(long double percent, int a, int b) {
	long double ans = 1.0L;
	percent /= 100.0L;
	for (int i = 0; i < a; ++i) {
		if (i < b) {
			ans *= percent;
		}
		else {
			ans *= (1.0L - percent);
		}
	}
	ans *= comb(a, b);
	return ans;
}

void solve(){
	int n, m, l;
	cin >> n >> m >> l;
	vector<tuple<int, int, long double>> inputs;
	REP(i, n) {
		int a, b, c;
		cin >> a >> b >> c;
		inputs.push_back(make_tuple(a, b, (long double)l / (long double)c));
	}
	REP(i, n) {
		long double ans = 0;
		for (int q = 0; q <= m; ++q) {
			long double req_time = q * get<1>(inputs[i]);
			req_time += get<2>(inputs[i]);
			long double pre = prob(get<0>(inputs[i]), m, q);
			REP(j, n) {
				if (j == i) continue;
				long double tmp = 0.0L;
				for (int t = 0; t <= m; ++t) {
					long double now_time = t * get<1>(inputs[j]);
					now_time += get<2>(inputs[j]);
					if (now_time > req_time + eps) {
						tmp += prob(get<0>(inputs[j]), m, t);
					}
				}
				pre *= tmp;
			}
			ans += pre;
		}
		cout << ans << endl;
	}
}

#undef int

int main() {
	init();
	solve();
}