/// Algoritmul lui Pepe

#include <cstdio>
#include <vector>
#include <deque>

using namespace std;

__attribute__((always_inline)) void read(int &num)
{
    static char inBuffer[0x30D40];

    static unsigned int p = 0x30D3F; num = 0x0;

    while(inBuffer[p] < 0x30 | inBuffer[p] > 0x39)
    {
        ++p != 0x30D40 || (fread(inBuffer, 0x1, 0x30D40, stdin), p = 0x0);
    }

    while(inBuffer[p] > 0x2F & inBuffer[p] < 0x3A)
    {
        num = num * 0xA + inBuffer[p] - 0x30;

        ++p != 0x30D40 || (fread(inBuffer, 0x1, 0x30D40, stdin), p = 0x0);
    }
}

char outBuffer[0x61A80]; unsigned int p;

__attribute__((always_inline)) void write(unsigned int x)
{
    unsigned int digits = x > 0x3B9AC9FF ? 0xA :
                 x > 0x5F5E0FF  ? 0x9 :
                 x > 0x98967F   ? 0x8 :
                 x > 0xF423F    ? 0x7 :
                 x > 0x1869F    ? 0x6 :
                 x > 0x270F     ? 0x5 :
                 x > 0x3E7      ? 0x4 :
                 x > 0x63       ? 0x3 :
                 x > 0x9        ? 0x2 : 0x1;

    for(unsigned int i = ~-digits; ~i; --i)
    {
        outBuffer[p + i] = x % 0xA + 0x30;

        x = x / 0xA;
    }

    p = p + digits; outBuffer[p++] = 0x20;
}

const int N=50000+7;
const int inf=(int)1e9;
int n,m;

struct edge
{
        int to;
        int cost;
};

vector <edge> g[N];
int d[N];
int mt[N];

int main()
{
        freopen("dijkstra.in","r",stdin);
        freopen("dijkstra.out","w",stdout);

        read(n);
        read(m);
        for(int i=2;i<=n;i++)
                d[i]=inf;
        for(int i=1;i<=m;i++)
        {
                int a,b,c;
                read(a);
                read(b);
                read(c);
                g[a].push_back({b,c});
        }

        for(int j=1;j<=n;j++)
                mt[j]=2;

        deque <int> dq;
        dq.push_back(1);
        while(!dq.empty())
        {
                int nod=dq.front();
                dq.pop_front();
                mt[nod]=0;
                for(auto &e : g[nod])
                        if(d[e.to]>d[nod]+e.cost)
                        {
                                d[e.to]=d[nod]+e.cost;
                                if(mt[e.to]==2)
                                        dq.push_back(e.to);
                                if(mt[e.to]==0)
                                        dq.push_front(e.to);
                                mt[e.to]=1;
                        }
        }

        for(int i=2;i<=n;i++)
        {
                if(d[i]==inf)
                        d[i]=0;
                write(d[i]);
        }
        puts(outBuffer);

        return 0;
}