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 — ui, vi, wi — the numbers of vertices connected by an edge and the weight of the edge (ui ≠ vi, 1 ≤ wi ≤ 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.

Examples

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.

题意：在满足最短路的情况下，使最短路的边的权值最小~输出权值，和边的序号~

题解：跑一遍最短路，在跑最短路的时候，判断权值最小，权值小只需要判断：在最短路相等的情况下，连接该点的边，找最小的那一个，这样就会使得权值最小~~上代码：

#include <iostream>
#include <cstring>
#include <queue>
using namespace std;
typedef long long ll;
const int inf=0x3f3f3f3f;
const int MAX = 1e6+100;
struct hh{
	int v,id,u;
	ll w;
	int nt;
	hh(){}//无参构造要写上
	hh (ll ww,int vv){//注意：括号里面的变量，要与结构体里面定义的变量不同
		w=ww;
		v=vv;
	}
	bool operator <(const hh &q) const{//重载小于号要写上，就是排序方式
		return w>q.w;//这里大于即小于，小于即大于，记住啊啊啊，老是忘了~~（算是重载小于号的性质吧）~~
	}
}a[MAX];
ll dis[MAX];
int tot,bian[MAX],head[MAX],len[MAX];
void add(int u,int v,int id,int w){
	a[tot].u=u;
	a[tot].v=v;
	a[tot].w=w;
	a[tot].nt=head[u];
	a[tot].id=id;
	head[u]=tot++;
}
void init(){
	memset(head,-1,sizeof(head));
	memset(dis,inf,sizeof(dis));
	memset(len,inf,sizeof(len));
}
void dij(int s){
	priority_queue<hh> q;
	dis[s]=0;
	len[s]=0;
	q.push(hh(dis[s],s));
	while(!q.empty()){
		hh tmp;
		tmp=q.top();
		q.pop();
		int u=tmp.v;
		for (int i = head[u];~i;i=a[i].nt){
			int v=a[i].v;
			if(dis[v]>dis[u]+a[i].w||(dis[v]==dis[u]+a[i].w&&len[v]>a[i].w)){
				dis[v]=dis[u]+a[i].w;//找最短路
				len[v]=a[i].w;//找到满足路长度相等的情况下，连接该点的最小边，即：或后面的判断
				bian[v]=a[i].id;//顶点对应的边，保存起来输出
				q.push(hh(dis[v],v));
			}
		}
	}
}
int main(){
	init();
	int n,m;
	cin >> n >> m;
	for (int i = 1; i <= m;i++){
		int u,v;
		ll w;
		cin >> u >> v >> w;
		add(u,v,i,w);
		add(v,u,i,w);
	}
	int x;
	cin >> x;
	dij(x);
	ll sum=0;
	for (int i = 1; i <= n;i++){
		sum+=len[i];//计算权值和
	}
	cout << sum << endl;
	bool ok=false;
	for (int i = 1; i <= n;i++){
		if(i!=x){//源点没有对应边~~
			if(ok){
				cout << " " << bian[i]; 
			}
			else{
				cout << bian[i];
				ok^=1;
			}
		}
	}
	cout << endl;
	return 0;
}