CF721C. Journey[DP DAG]

C. Journey

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Recently Irina arrived to one of the most famous cities of Berland — the Berlatov city. There are n showplaces in the city, numbered from 1 to n, and some of them are connected by one-directional roads. The roads in Berlatov are designed in a way such that there are nocyclic routes between showplaces.

Initially Irina stands at the showplace 1, and the endpoint of her journey is the showplace n. Naturally, Irina wants to visit as much showplaces as she can during her journey. However, Irina's stay in Berlatov is limited and she can't be there for more than T time units.

Help Irina determine how many showplaces she may visit during her journey from showplace 1 to showplace n within a time not exceeding T. It is guaranteed that there is at least one route from showplace 1 to showplace n such that Irina will spend no more than Ttime units passing it.

Input

The first line of the input contains three integers n, m and T (2 ≤ n ≤ 5000,  1 ≤ m ≤ 5000,  1 ≤ T ≤ 109) — the number of showplaces, the number of roads between them and the time of Irina's stay in Berlatov respectively.

The next m lines describes roads in Berlatov. i-th of them contains 3 integers ui, vi, ti (1 ≤ ui, vi ≤ n, ui ≠ vi, 1 ≤ ti ≤ 109), meaning that there is a road starting from showplace ui and leading to showplace vi, and Irina spends ti time units to pass it. It is guaranteed that the roads do not form cyclic routes.

It is guaranteed, that there is at most one road between each pair of showplaces.

Output

Print the single integer k (2 ≤ k ≤ n) — the maximum number of showplaces that Irina can visit during her journey from showplace 1 to showplace n within time not exceeding T, in the first line.

Print k distinct integers in the second line — indices of showplaces that Irina will visit on her route, in the order of encountering them.

If there are multiple answers, print any of them.

Examples

input

4 3 13
1 2 5
2 3 7
2 4 8

output

3
1 2 4 

input

6 6 7
1 2 2
1 3 3
3 6 3
2 4 2
4 6 2
6 5 1

output

4
1 2 4 6 

input

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

output

3
1 3 5 

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题意：单向，没有回路，没有重边自环，限制时间，求1到n最多经过几个点，并输出这些点任意方案

---

因为没有环，又保证1和n连通，一开始想树形DP，并不好做，然后发现这是有向边
突然发现，这不就是有向无环图，有向无环图DAG的最短路最长路可以用DP来做，扩展一下应该也可以
f[i][j]表示从i到n经过j个点的时间

PS：因为忘判vis TLE一次

//
//  main.cpp
//  c
//
//  Created by Candy on 9/30/16.
//  Copyright © 2016 Candy. All rights reserved.
//

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <vector>
#include <string>
using namespace std;
const int N=,M=,INF=1e9+;
inline int read(){
    char c=getchar();int x=,f=;
    while(c<''||c>''){if(c=='-')f=-;c=getchar();}
    while(c>=''&&c<=''){x=x*+c-'';c=getchar();}
    return x;
}
int n,m,T,u,v,w;
struct edge{
    int v,w,ne;
}e[M<<];
int h[N],cnt=;
void ins(int u,int v,int w){
    cnt++;
    e[cnt].v=v;e[cnt].w=w;e[cnt].ne=h[u];h[u]=cnt;
}
int f[N][N],vis[N];
void dp(int u){
    if(u==n) return;
    int child=;
    if(vis[u]) return;
    vis[u]=;
    for(int i=h[u];i;i=e[i].ne){
        child++;
        int v=e[i].v,w=e[i].w;
        dp(v);
        for(int j=;j<=n;j++) if(f[v][j-]<INF)
            f[u][j]=min(f[u][j],f[v][j-]+w);
    }
}
void print(int u,int d){
    printf("%d ",u);
    for(int i=h[u];i;i=e[i].ne){
        int v=e[i].v,w=e[i].w;
        if(f[v][d-]<INF&&f[u][d]==f[v][d-]+w) {print(v,d-);break;}
    }
}
int main(int argc, const char * argv[]) {
    n=read();m=read();T=read();
    for(int i=;i<=m;i++){
        u=read();v=read();w=read();
        ins(u,v,w);
    }
    memset(f,,sizeof(f));
    f[n][]=;
    dp();
    int num=;
    for(int i=n;i>=;i--)
        if(f[][i]<=T) {num=i;break;}
    printf("%d\n",num);
    print(,num);
    return ;
}