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 *G*<sub class="lower-index">1</sub> = (*V*, *E*<sub class="lower-index">1</sub>) that is a tree with the set of edges *E*<sub class="lower-index">1</sub> 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 *G*<sub class="lower-index">1</sub> 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.
The first line contains two numbers, *n* and *m* (1 ≤ *n* ≤ 3·10<sup class="upper-index">5</sup>, 0 ≤ *m* ≤ 3·10<sup class="upper-index">5</sup>) — the number of vertices and edges of the graph, respectively.

Next *m* lines contain three integers each, representing an edge — *u*<sub class="lower-index">*i*</sub>, *v*<sub class="lower-index">*i*</sub>, *w*<sub class="lower-index">*i*</sub> — the numbers of vertices connected by an edge and the weight of the edge (*u*<sub class="lower-index">*i*</sub> ≠ *v*<sub class="lower-index">*i*</sub>, 1 ≤ *w*<sub class="lower-index">*i*</sub> ≤ 10<sup class="upper-index">9</sup>). 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.
import heapq

n, m = map(int, input().split())
g = [[] for _ in range(n + 1)]
for i in range(1, m + 1):
    u, v, w = map(int, input().split())
    g[u].append((i, v, w))
    g[v].append((i, u, w))
src = int(input())

pq = [(0, 0, src, -1)]
mk = [0] * (n + 1)
t = []
s = 0
while pq:
    d, w, u, e = heapq.heappop(pq)
    if mk[u]:
        continue
    mk[u] = 1
    s += w
    t.append(e)
    for e, v, w in g[u]:
        if not mk[v]:
            heapq.heappush(pq, (d + w, w, v, e))

print(s)
print(*t[1:])