/*
此题纯属套模板
调用时,初始结点s,目标结点e,则
SPFA(s);
cout<<d[e]<<endl;
即可。注意结点是从1存储到N。
不能连通时值为MAX*/
#include <stdio.h>
#include <string.h>
int d[1002],n,m;
int edges[1005][1005];
int queue[1000001];
#define MAX 999999999
#define N 1001
/*
int SPFA(int s)
{
int i;
bool visit[N] = {false};
int front = 0, rear = 1;
memset(queue,0,sizeof(queue));
for(i=1;i<=n;i++)
d[i] = MAX;
//int path[N];
queue[front] = s;
visit[s] = true;
d[s] = 0;
while(front<rear)
{
int u = queue[front];
visit[u] = false;
for(int i=1; i<=n; i++)
{
if (d[i]>d[u]+ edges[u][i])
{
d[i]= d[u]+edges[u][i];
//path[i] = u;
if( !visit[i] )
{
visit[i] = true;
queue[rear++] = i;
}
}
}
front++;
}
return 0;
}
*/
void dijkstra(int v)
{
int i,j;
bool s[N]={false};
for(i=1;i<=n;i++)
d[i]=edges[v][i];
d[v]=0;s[v]=true;
for(i=1;i<n;i++)
{
int temp=MAX;
int u=v;
for(j=1;j<=n;j++)
if((!s[j])&&(d[j]<temp))
{
u=j;
temp=d[j];
}
s[u]=true;
for(j=1;j<=n;j++)
if((!s[j])&&(edges[u][j]<MAX)&&(d[u]+edges[u][j])<d[j])
d[j]=d[u]+edges[u][j];
}
}
int main()
{
int i,j,a,b,c;
while(scanf("%d%d",&n,&m)!=EOF&&(n||m))
{
for(i=0;i<=1000;i++){
for(j=0;j<=i;j++)
{
edges[i][j]=MAX;
edges[j][i]=MAX;
}
}
for(i=0;i<m;i++)
{
scanf("%d %d %d",&a,&b,&c);
if(edges[a][b]>c){
edges[a][b]=c;
edges[b][a]=c;
}
}
// SPFA(1);
dijkstra(1);
printf("%d\n",d[n]);
}
return 0;
}