只是想看看哪里RE了。

#include<bits/stdc++.h>
#define sd std::
#define int long long
#define inf 0x3f3f3f3f
#define linf 1e18
#define il inline
#define db double
#define ldb long double
#define F(i,a,b) for(int i=(a);i<=(b);i++)
#define f(i,a,b) for(int i=(a);i>=(b);i--)
#define MIN(x,y) (x<y?x:y)
#define MAX(x,y) (x>y?x:y)
#define me(x,y) memset(x,y,sizeof x)
#define pii sd pair<int,int>
#define umap(x,y) sd unordered_map<x,y>
#define pque(x) sd priority_queue<x>
#define X first
#define Y second
#define kg putchar(' ')
#define Fr(a) for(auto it:a)
#define dbg(x) sd cout<<#x<<": "<<x<<sd endl
il int read(){int w=1,c=0;char ch=getchar();for(;ch>'9'||ch<'0';ch=getchar()) if(ch=='-') w=-1;for(;ch>='0'&&ch<='9';ch=getchar()) c=(c<<3)+(c<<1)+ch-48;return w*c;}
void printt(int x){if(x>9) printt(x/10);putchar(x%10+48);}
il void print(int x){if(x<0) putchar('-'),printt(-x);else printt(x);}
il void put(int x){print(x);putchar('\n');}
il void printk(int x){print(x);kg;}
const int N=1e5+100;
const int mod=1e9+7;
int po[N],h[N];//预处理2次幂以及hash值 
int n,m,ma;//ma记录最大边权 
struct node
{
	int nex;
	int to;
	int w;
}a[N<<1];
int cnt,head[N];
void add(int u,int v,int w)
{
	a[++cnt].nex=head[u];
	head[u]=cnt;
	a[cnt].to=v;
	a[cnt].w=w;
}
struct pst_tree
{
	struct node
	{
		int l,r;
		int val;
		int ha;//哈希值 
		bool pd;//记录区间是否全为1，用来找最长1。 
	}t[N<<7];
	int root[N],tot=0;
	#define l(i) t[i].l
	#define r(i) t[i].r
	#define val(i) t[i].val
	#define ha(i) t[i].ha
	#define pd(i) t[i].pd
	void pushup(int k)
	{
		pd(k)=pd(l(k))&pd(r(k));
		val(k)=(val(l(k))+val(r(k)))%mod;
		ha(k)=(ha(l(k))+ha(r(k)))%mod;
	}
	int newly(int k)
	{
		t[++tot]=t[k];
		return tot;
	}
	void built(int &k,int l,int r,int v)
	{
		k=++tot;
		if(l==r) return val(k)=po[l]*v,ha(k)=h[l]*v,pd(k)=v,void();
		int mid=l+r>>1;
		built(l(k),l,mid,v);
		built(r(k),mid+1,r,v);
		pushup(k);
	}
	bool check(int k,int l,int r,int x,int y)//判区间是否为1 
	{
		if(!k) return 0;
		if(x<=l&&y>=r) return pd(k);
		int mid=l+r>>1,sum=true;
		if(x<=mid) sum&=check(l(k),l,mid,x,y);
		if(y>mid) sum&=check(r(k),mid+1,r,x,y);
		return sum;
	}
	void upd_dan(int &k,int l,int r,int x)//单点修1 
	{
		k=newly(k);
		if(l==r) return val(k)=po[l],ha(k)=h[l],pd(k)=1,void();
		int mid=l+r>>1;
		if(x<=mid) upd_dan(l(k),l,mid,x);
		else upd_dan(r(k),mid+1,r,x);
		pushup(k);
	}
	void upd_qu(int &k,int l,int r,int x,int y)//区间修0
	{
		k=newly(k);
		if(x<=l&&y>=r) return k=0,void();
		int mid=l+r>>1;
		if(x<=mid) upd_qu(l(k),l,mid,x,y);
		if(y>mid) upd_qu(r(k),mid+1,r,x,y);
		pushup(k);
	}
	void add(int u,int v)
	{
		int l=v,r=ma,ans=v-1;
		while(l<=r)
		{
			int mid=l+r>>1;
			if(check(root[u],0,ma,v,mid)) ans=mid,l=mid+1;
			else r=mid-1;
		}
		root[n+1]=root[u];
		if(v<=ans) upd_qu(root[n+1],0,ma,v,ans);
		upd_dan(root[n+1],0,ma,ans+1);
	}
	bool cmp(int x,int y,int l,int r)//两颗主席树比大小,比较区间为l~r 
	{
		if(l==r) return pd(x)>pd(y);
		int mid=l+r>>1;
		if(val(r(x))!=val(r(y))&&ha(r(x))!=ha(r(y))) return cmp(r(x),r(y),mid+1,r);
		else return cmp(l(x),l(y),l,mid);
	}
}T;
struct dcz
{
    int u,d;
    bool operator<(const dcz& b)const
	{
        return T.cmp(d,b.d,0,ma);
    }
};
int st,en;//起点终点 
int vis[N],pre[N],ans[N];
pque(dcz) q;
void dij()
{
	T.built(T.root[st],0,ma,0);
	T.built(T.root[0],0,ma,1);
	F(i,1,n)
	{
		if(i==st) continue;
		T.root[i]=T.root[0];
	}
	q.push({st,T.root[st]});
    while(!q.empty())
	{
        auto [u,d]=q.top();
        q.pop();
//        put(u);
        if(vis[u]) continue;
        vis[u]=1;
        if(u==en) return;
        for(int i=head[u];i;i=a[i].nex)
		{
            int v=a[i].to;
            T.add(u,a[i].w);
            if(T.cmp(T.root[v],T.root[n+1],0,ma))
            {
            	T.root[v]=T.root[n+1],pre[v]=u;//顺便记录前驱 
            	if(!vis[v]) q.push({v,T.root[v]});
            }
        }
    }
}
il void init()
{
	po[0]=1,h[0]=1;
	F(i,1,ma) po[i]=(po[i-1]<<1)%mod,h[i]=(h[i-1]*123)%mod;
}
il void solve()
{
	n=read(),m=read();
	F(i,1,m)
	{
		int x=read(),y=read(),z=read();
		ma=MAX(ma,z)+100;
		add(x,y,z);
		add(y,x,z);
	}
	st=read(),en=read();
	init();
	dij();
	if(!pre[en]) return put(-1);
	put(T.val(T.root[en]));
	int c=0;
	while(en!=st) ans[++c]=en,en=pre[en];//倒退回去 
	put(c+1);
	printk(st);
	f(i,c,1) printk(ans[i]);
	return;
}
signed main()
{
	int T=1;
//	T=read();
	while(T--) solve();
    return 0;
}