// P2047 社交网络。正权、无向、简单、连通图；有序点对，排除端点。
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
int main() {
    ios::sync_with_stdio(false); cin.tie(nullptr);
    int n, m; if (!(cin >> n >> m)) return 0;
    const ll INF = 1LL << 60;
    vector<vector<ll>> d(n, vector<ll>(n, INF));
    // 中间受限图的计数可能超过最终答案范围，从存储起即用浮点。
    vector<vector<long double>> cnt(n, vector<long double>(n, 0));
    for (int i = 0; i < n; ++i) d[i][i] = 0;
    for (int e = 0, u, v; e < m; ++e) {
        ll w; cin >> u >> v >> w; --u; --v;
        d[u][v] = d[v][u] = w; cnt[u][v] = cnt[v][u] = 1;
    }
    for (int k = 0; k < n; ++k) for (int i = 0; i < n; ++i) for (int j = 0; j < n; ++j) {
        if (i == k || j == k || i == j) continue;
        if (d[i][k] == INF || d[k][j] == INF) continue;
        ll nd = d[i][k]+d[k][j];
        long double ways = cnt[i][k]*cnt[k][j];
        if (nd < d[i][j]) { d[i][j] = nd; cnt[i][j] = ways; }
        else if (nd == d[i][j]) cnt[i][j] += ways;
    }
    cout << fixed << setprecision(3);
    for (int v = 0; v < n; ++v) {
        long double ans = 0;
        for (int s = 0; s < n; ++s) for (int t = 0; t < n; ++t) {
            if (s == t || s == v || t == v) continue;
            if (d[s][v]+d[v][t] == d[s][t]) ans += cnt[s][v]*cnt[v][t]/cnt[s][t];
        }
        cout << ans << '\n';
    }
}