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 <cstdio>
#include <vector>
#include <cmath>
#include <algorithm>
#include <cassert>

using namespace std;

typedef long double real;
#define EQ(x,y) (abs((x)-(y))<EPS)

real EPS = 1e-9;

const int MAX_N = 101;
const int MAX_M = 53;

const bool debug = false;

int N, M; real L;
vector<real> P, T, V;
void input() {
    cin >> N >> M >> L;
    P.clear(); P.resize(N);
    T.clear(); T.resize(N);
    V.clear(); V.resize(N);
    for (int i = 0; i < N; i++) {
        cin >> P[i] >> T[i] >> V[i];
        P[i] /= 100;
    }
}

real X[MAX_N][MAX_M]; // X[i][j] :: i人目がm回休む確率
real S[MAX_N][MAX_M]; // S[i][j] :: i人目がm回未満だけ休む確率

void init() {
    for (int i = 0; i < N; i++) for (int j = 0; j <= M; j++) X[i][j] = 0;
    for (int i = 0; i < N; i++) for (int j = 0; j <= M + 1; j++) S[i][j] = 0;
    for (int i = 0; i < N; i++) {
        static real dp[MAX_M][MAX_M];
        for (int j = 0; j <= M; j++) for (int k = 0; k <= M; k++) dp[j][k] = 0;
        dp[0][0] = 1;
        for (int j = 0; j < M; j++) {
            for (int k = 0; k <= j; k++) {
                dp[j + 1][k] += dp[j][k] * (1.0 - P[i]);
                dp[j + 1][k + 1] += dp[j][k] * P[i];
            }
        }
        for (int k = 0; k <= M; k++) {
            X[i][k] = dp[M][k];
        }
    }
    for (int i = 0; i < N; i++) {
        S[i][0] = 0;
        for (int j = 0; j <= M; j++) {
            S[i][j + 1] = S[i][j] + X[i][j];
        }
    }
}

void solve() {
    init();
    for (int i = 0; i < N; i++) { // i人目
        if (V[i] == 0) {
            printf("%.12Lf\n", real(0));
            continue;
        }
        real ans = 0;
        for (int j = 0; j <= M; j++) { // j回休んだ
            real p = X[i][j];
            real time = L / V[i] + j * T[i];
            for (int k = 0; k < N; k++) { // k人目
                if (i == k) continue;
                if (EQ(T[k], 0)) {
                    p *= (L / V[k] - EPS > time);
                } else {
                    // c回以上休む必要
                    int c = max(0, int(ceil((time - L / V[k]) / T[k] + EPS) + EPS));
                    if (c > M) {
                        p = 0;
                        break;
                    }
                    real pk = 1.0 - S[k][c];
                    p *= pk;
                }
            }
            ans += p;
        }
        printf("%.12Lf\n", ans);
    }
}

int main() {
    input(); solve();
    return 0;
}