#include <iostream>
#include <iomanip>
#include <vector>
#include <algorithm>
#include <functional>
#include <cmath>
#include <string>
#include <queue>
#include <map>
#include <bitset>
#include <set>
#include <stack>
#include <numeric>
#include <unordered_map>
#include <random>

using namespace std;

using ll = long long;
using vi = vector<int>;
using vvi = vector<vi>;
using vl = vector<ll>;
using vvl = vector<vl>;
using vb = vector<bool>;
using vvb = vector<vb>;
using vd = vector<double>;
using vs = vector<string>;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using pdd = pair<double, double>;
using vpii = vector<pii>;
using vpll = vector<pll>;
using vpdd = vector<pdd>;
const int inf = (1 << 30) - 1;
const ll INF = 1LL << 60;
//const int MOD = 1000000007;
const int MOD = 998244353;

struct Edge {
	int to;
	ll cost;
};

using Graph = vector<vector<Edge>>;
vl memo;
ll dfs(int s, Graph& g) {
	if (memo[s]) return memo[s];

	ll ret = 0;
	
	for (auto& v : g[s]) {
		ll x = dfs(v.to, g);
		ret += x * v.cost;
	}
	return memo[s] = ret;
}

int main() {
	int n, m;
	cin >> n >> m;
	vi p(m), q(m), r(m);
	for (int i = 0; i < m; i++) {
		cin >> p[i] >> q[i] >> r[i];
	}
	vl ans(n + 1, 0);
	Graph g(n + 1);
	// 辺を逆向きにして製品を作るのに必要な材料の数をDFSで求める方法はTLE
	// 大元の材料が次の材料に使われる数をメモ化再帰で求める
	for (int i = 0; i < m; i++) {
		g[p[i]].push_back({ r[i], q[i] });
	}

	// 頂点の入次数を求める
	vi indeg(n + 1, 0);
	for (int i = 0; i < m; i++) {
		indeg[r[i]]++;
	}
	// memo[i]: 材料iが必要な数を保存
	// 頂点N = 1 から逆向きに決まる
	memo.assign(n + 1, 0);
	memo[n] = 1;

	// 入次数 0 の頂点からDFSする
	for (int i = 1; i < n; i++) {
		if (indeg[i] == 0) dfs(i, g);
	}

	for (int i = 1; i < n; i++) {
		if (indeg[i] == 0) cout << memo[i] << endl;
		else cout << 0 << endl;
	}

	return 0;
}