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#include <iostream>
#include <fstream>
#include <vector>
#include <map>
#include <queue>
using namespace std;
ifstream fin("dijkstra.in");
ofstream fout("dijkstra.out");
const int MAX_SIZE = 50005;
const int MAX_SUM = 20000000;
vector<int> graph[MAX_SIZE];
int costs[MAX_SIZE][MAX_SIZE];
int minSum = MAX_SUM, currentSum = 0;
int prevSum = 0, curSum = 0;
void findMinimumCost(int endPeak, int currentPeak, int prevPeak) {
if (graph[currentPeak].size() == 0 || currentPeak == endPeak) {
if (currentPeak == endPeak && currentSum < minSum) {
minSum = currentSum;
}
//cout << "in return:" << prevPeak << " " << currentPeak << " " << currentSum << "\n";
prevSum = prevPeak;
curSum = currentPeak;
return;
} // trebuie sa vad cum schimb valoarea lui currentSum cand ma intorc din recursivitate
for (int i = 0; i < graph[currentPeak].size(); ++i) {
currentSum += costs[currentPeak][graph[currentPeak][i]];
//cout << currentPeak << " " << graph[currentPeak][i] << " " << currentSum << "\n";
findMinimumCost(endPeak, graph[currentPeak][i], currentPeak);
//cout << "ne intoarcem: ";
currentSum -= costs[prevSum][curSum];
prevSum = prevPeak;
curSum = currentPeak;
//if (prevPeak == 0) {
// currentSum = 0;
//}
//cout << prevPeak << " " << currentPeak << " " << currentSum << "\n";
//cout << "\n";
}
}
int main() {
int peaks, arches;
fin >> peaks >> arches;
for (int i = 1; i <= arches; ++i) {
int start, end, cost;
fin >> start >> end >> cost;
graph[start].push_back(end);
costs[start][end] = cost;
}
for (int destination = 2; destination <= peaks; ++destination) {
minSum = MAX_SUM;
findMinimumCost(destination, 1, 0);
if (minSum == MAX_SUM) {
minSum = 0;
}
fout << minSum << " ";
}
return 0;
}
/*
1 ≤ N ≤ 50 000
1 ≤ M ≤ 250 000
Lungimile arcelor sunt numere naturale cel mult egale cu 20 000.
Pot exista arce de lungime 0
Nu exista un arc de la un nod la acelasi nod.
Daca nu exista drum de la nodul 1 la un nod i, se considera ca lungimea minima este 0.
Arcele date in fisierul de intrare nu se repeta.
5 7
1 2 1
2 3 2
3 4 3
4 5 1
1 5 700
2 5 100
3 5 30 -> 1 3 6 7
5 3
1 5 2
1 4 5
5 4 1 -> 0 0 3 2
5 3
1 5 2
1 4 3
5 4 1 -> 0 0 3 2
3 1
1 2 1 -> 1 0
4 4
1 2 1
1 3 2
3 4 1
2 4 1 -> 1 2 2
5 6
1 2 1
1 4 2
4 3 4
2 3 2
4 5 3
3 5 6 -> 1 3 2 5
*/