#include<bits/stdc++.h>#define F(i,a,b) for (int i=a;i<b;i++)#define FF(i,a,b) for (int i=a;i<=b;i++)#define mes(a,b) memset(a,b,sizeof(a))#define INF 0x3f3f3f3f#define LL long longusing namespace std;const int N=5100,MAX=1000100;int n,m,T,dp[N][N],pre[N][N];   //dp[i][j]表示走过j个点到达第i个点的时间，pre记录父亲路径int w[N],head[N],from[N],to[N],nexte[N];void AddEdge(int u,int v,int t,int e){    from[e]=u;    to[e]=v;    nexte[e]=head[u];    head[u]=e;    w[e]=t;}void PrintPath(){    int num;    FF(i,1,n){        if(dp[n][i]<=T)num=i;    }printf("%d\n",num);    stack<int >S;    int k=n;    for(int i=num;i>0;i--){        S.push(k);        k=pre[k][i];    }    printf("%d",S.top());S.pop();    while(!S.empty()){        printf(" %d",S.top());        S.pop();    }puts("");}void Solve(){    mes(dp,INF);    dp[1][1]=0;    FF(i,2,n){      //直接双层循环搞出dp[][]的值        FF(j,1,m){            int u=from[j],v=to[j],val=w[j];            if(dp[u][i-1]+val<=T&&dp[u][i-1]+val<dp[v][i]){                dp[v][i]=dp[u][i-1]+val;                pre[v][i]=u;            }        }    }    PrintPath();}int main(){    while(~scanf("%d%d%d",&n,&m,&T))    {        mes(w,INF);        mes(head,-1);        int u,v,t;        FF(i,1,m){            scanf("%d%d%d",&u,&v,&t);            AddEdge(u,v,t,i);        }        Solve();    }    return 0;}