E. Paths and Trees

Little girl Susie accidentally found her elder brother's notebook. She has many things to do, more important than solving problems, but she found this problem too interesting, so she wanted to know its solution and decided to ask you about it. So, the problem statement is as follows.

Let's assume that we are given a connected

weighted undirected graph G = (V, E) (here V is the set of vertices, E is the set of edges). The shortest-path tree from vertex u is such graph G₁ = (V, E₁) that is a tree with the set of edges E₁ that is the subset of the set of edges of the initial graph E, and the lengths of the shortest paths from u

to any vertex to G and to G₁ are the same.

You are given a connected weighted undirected graph G and vertex u. Your task is to find the shortest-path tree of the given graph from vertex u, the total weight of whose edges is minimum possible.

Input

The first line contains two numbers, n and m (1 ≤ n

 ≤ 3·10⁵, 0 ≤ m ≤ 3·10⁵) — the number of vertices and edges of the graph, respectively.

Next m lines contain three integers each, representing an edge — u_i, v_i, w_i — the numbers of vertices connected by an edge and the weight of the edge (u_i ≠ v_i, 1 ≤ w_i ≤ 10⁹). It is guaranteed that graph is connected and that there is no more than one edge between any pair of vertices.

The last line of the input contains integer u (1 ≤ u ≤ n) — the number of the start vertex.

Output

In the first line print the minimum total weight of the edges of the tree.

In the next line print the indices of the edges that are included in the tree, separated by spaces. The edges are numbered starting from 1 in the order they follow in the input. You may print the numbers of the edges in any order.

If there are multiple answers, print any of them.

Sample test(s) Input

3 3
1 2 1
2 3 1
1 3 2
3

Output

2
1 2 

Input

4 4
1 2 1
2 3 1
3 4 1
4 1 2
4

Output

4
2 3 4 

Note

In the first sample there are two possible shortest path trees:

• with edges 1 – 3 and 2 – 3 (the total weight is 3);
• with edges 1 – 2 and 2 – 3 (the total weight is 2);

And, for example, a tree with edges 1 – 2 and 1 – 3 won't be a shortest path tree for vertex 3, because the distance from vertex 3 to vertex 2 in this tree equals 3, and in the original graph it is 1.

題意：用最短路構造最小生成樹，輸出總的權值是多少，用到了那些邊

思路：SPFA求出單源最短路，並記下用到了那些邊，然後統計輸出答案，值得注意的是，最短路有多條的時候

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int maxn=300010;
const LL INF=1e16;
struct E
{
    int v,next,w,id;
}edge[maxn*2];
int N,M;
int st;
LL dis[maxn];
bool vis[maxn];
int W[maxn];
int head[maxn],tot;
int pre[maxn],cnt[maxn];
void init()
{
    tot=0;
    memset(head,-1,sizeof(head));
    memset(pre,0,sizeof(pre));
    memset(cnt,0,sizeof(cnt));
}
void add_edge(int u,int v,int w,int id)
{
    edge[tot].v=v;
    edge[tot].next=head[u];
    edge[tot].w=w;
    edge[tot].id=id;
    head[u]=tot++;
}
void SPFA()
{
    for(int i=0;i<=N;i++)
        dis[i]=INF,vis[i]=0;
    queue<int> q;
    q.push(st);
    dis[st]=0;
    vis[st]=1;
    while(!q.empty())
    {
        int u=q.front();q.pop();
        vis[u]=0;
        for(int i=head[u];i!=-1;i=edge[i].next)
        {
            int v=edge[i].v;
            int w=edge[i].w;
            if(dis[v]>dis[u]+w)
            {
                dis[v]=dis[u]+w;
                pre[v]=edge[i].id;
                //pre[v]=make_pair(u,edge[i].id);
                if(!vis[v])
                {
                    q.push(v);
                    vis[v]=1;
                }
            }
            else if(dis[v]==dis[u]+w)
            {
                if(w<W[pre[v]])
                {
                    pre[v]=edge[i].id;
                    if(!vis[v])
                    {
                        q.push(v);
                        vis[v]=1;
                    }
                }
            }
        }
    }

}

int main()
{
    scanf("%d%d",&N,&M);
    init();
    for(int i=1;i<=M;i++)
    {
        int u,v,w;
        scanf("%d%d%d",&u,&v,&w);
        add_edge(u,v,w,i);
        add_edge(v,u,w,i);
        W[i]=w;
    }
    scanf("%d",&st);
    SPFA();
    memset(vis,0,sizeof(vis));
    for(int i=1;i<=N;i++)
        vis[pre[i]]=1;
    LL sum=0;
    for(int i=1;i<=N;i++)
        if(pre[i]>0)sum+=W[pre[i]];
    cout<<sum<<endl;
    for(int i=1;i<=N;i++)
        if(pre[i])printf("%d ",pre[i]);
    printf("\n");
    return 0;
}