You are given a weighted undirected graph. The vertices are enumerated from 1 to n. Your task is to find the shortest path between the vertex 1 and the vertex n.

Input

The first line contains two integers n and m (2 ≤ n ≤ 105, 0 ≤ m ≤ 105), where n is the number of vertices and m

It is possible that the graph has loops and multiple edges between pair of vertices.

Output

Write the only integer -1 in case of no path. Write the shortest path in opposite case. If there are many solutions, print any of them.

Examples

input

5 6
1 2 2
2 5 5
2 3 4
1 4 1
4 3 3
3 5 1

output

1 4 3 5 

input

5 6
1 2 2
2 5 5
2 3 4
1 4 1
4 3 3
3 5 1

output

1 4 3 5 

#include <bits/stdc++.h>
#define INF 0x3fffffff
#define mem(ar,num) memset(ar,num,sizeof(ar))
#define me(ar) memset(ar,0,sizeof(ar))
#define lowbit(x) (x&(-x))
#define IOS ios::sync_with_stdio(false)
#define DEBUG cout<<endl<<"DEBUG"<<endl;
using namespace std;
int n,m,head[100100],a,b,c,cnt,flag;
long long dis[100100];int vis[100100],po[100100],ans[100100];
struct no{
    int next,to,w;
}edge[500100*2];
void add(int u,int v,int w){
    edge[++cnt].next=head[u];
    edge[cnt].to=v;
    edge[cnt].w=w;
    head[u]=cnt;
}
struct node{
    int dis;
    int pos;
    bool operator <( const node &x )const{
        return x.dis < dis;
    }
};
bool dijkstra(){
    priority_queue<node> q;
    dis[1]=0;
    ans[flag]=1;
    q.push((node){0,1});
    while(!q.empty()){
        node tmp = q.top();
        q.pop();
        int x = tmp.pos, d = tmp.dis;
        if(vis[x])
            continue;
        vis[x] = 1;
        int t=head[x];
        while(t){
            int y=edge[t].to;
            if(dis[y] > dis[x] + edge[t].w){
                dis[y] = dis[x] + edge[t].w;
                if(!vis[y]){
                    po[edge[t].to]=tmp.pos;//記錄路徑
                    q.push((node){dis[y],y});
                }
            }
            t=edge[t].next;
        }
    }
    if(dis[n]==0x7f7f7f7f7f7f7f7f)return false;
    return true;
}
int main(){
    mem(dis,0x7f);
    cin>>n>>m;
    for(int i=0;i<m;i++){
        cin>>a>>b>>c;
        add(a,b,c);
        add(b,a,c);
    }
    if(dijkstra()){
        for(int i=n;i;i=po[i]){
            ans[++flag]=i;
        }
        for(int i=flag;i>=1;i--)
            cout<<ans[i]<<" ";
    }else cout<<-1<<endl;
    return 0;
}