#include <bits/stdc++.h>
#define ll long long
#define nl '\n'
#define FOR(i, a, b) for (int i = a; i <= b; ++i) 
#define F0R(i, a, b) for (int i = a; i >= b; --i)
#define FORd(i, n) for (int i = 0; i < n; ++i)
#define F0Rd(i, n) for (int i = n - 1; i >= 0; --i)
#define trav(a, x) for (auto &a : x)
#define uid(a, b) uniform_int_distribution<int>(a, b)(rng)
using namespace std;

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

const int mxN = 1e5 + 5, INF = 1e9 + 7;

ifstream fin("bellmanford.in");
ofstream fout("bellmanford.out");

int N, M, dist[mxN], cnt_in_queue[mxN];
vector<pair<int, int>> edges[mxN];
bitset<mxN> inQueue;

void BF() {
    queue<int> q;
    q.push(1);
    dist[1] = 0;
    while (q.size()) {
        int node = q.front();
        q.pop();
        inQueue[node] = false;
        for (pair<int, int> elem : edges[node]) {
            int next_node = elem.first;
            int new_cost = elem.second;
            if (dist[next_node] > dist[node] + new_cost) {
                dist[next_node] = dist[node] + new_cost;
                if (cnt_in_queue[next_node] > N) {
                    fout << "Ciclu negativ!";
                    return;
                }
                q.push(next_node);
                inQueue[next_node] = true;
                cnt_in_queue[next_node]++;
            }
        }
    }
    for (int i = 2; i <= N; ++i) {
        fout << dist[i] << " ";
    }
}

void solve() {
    fin >> N >> M; 
    for (int i = 1; i <= N; ++i) {
        dist[i] = INF;
    }
    while (M--) {
        int x, y, z;
        fin >> x >> y >> z;
        edges[x].push_back({y, z});
    }
    BF();
}

int main() {
    fin.tie(0); fout.tie(0);
    ios::sync_with_stdio(0);
    int T = 1;
//    cin >> T;
    while (T--) {
        solve();
    }
    return 0;
}

//read the question correctly (ll vs int)
//what's the meaning of the problem ? Think outside the BOX !!!
//edge cases ?
//make it simple
//write everything (observations, edge cases, ideas, steps, methods, ...)