/**vector d( cu dist minime) in care toate inafara de primu is infinit oo=constanta mare aproape de infinit
luam minimu din d si ii relaxam vecinii (daca distanta panal la el + d[don curent]ii mai mica decat d[nodul resp le bagam])
vector bool s cu in care nod am mai fost sa nu repetam


pair <int,int>P[100];
P=makepair(1,2);
cout<<P.first<<P.second;*/
#include <iostream>
#include <fstream>
#include <vector>
using namespace std;
ifstream in("dijkstra.in");
ofstream out("dijkstra.out");

const int NMax=50000;
const int oo=2000000000;
int D[NMax+5],S[NMax+5],n,m,i,x,y,z,Nod;
vector <pair<int,int> > G[NMax+5];

void citire()
{
    in>>n>>m;
    for(int i=1;i<=n;++i)
    {
        in>>x>>y>>z;
        G[x].push_back(make_pair(y,z));
    }
}

void dijkstra()
{
    for(int i=2;i<=m;++i)
        D[i]=oo;

    for(int k=1;k<=m;++k)
    {
        int Min=oo;
        for(int i=1;i<=m;++i)
            if(D[i]<Min && S[i]==0)
            {
                Min=D[i];
                Nod=i;
            }
        S[Nod]=1;
        for(int i=0;i< (int)G[Nod].size();++i)
        {
            int Vecin=G[Nod][i].first;
            int Cost=G[Nod][i].second;
            D[Vecin]=min(D[Vecin],D[Nod]+Cost);
        }
    }
}

void afisare()
{
    for(i=2;i<=n;++i)
        out<<D[i]<<" ";
}

int main()
{
    citire();
    dijkstra();
    afisare();
}