<|instruction|>
Solve the below problem. The solution should start with an abridged problem statement. Then key observations. Then full solution based on the observations. Then C++ and Python implementations with comments.

515. Recover path
Time limit per test: 0.75 second(s)
Memory limit: 262144 kilobytes
input: standard
output: standard

Traveller Gregory is famous for his ability to always choose the shortest path for his journey. Ostap is a journalist who seeks for information about the recent Gregory's trip. He managed to get the evidence that during this trip Gregory visited a number of cities. However, there is no information about the order in which they were visited, and no information about the starting city and the ending city of Gregory's trip (the entire trip is (one of) the shortest path between these cities). Help Ostap to find any shortest path that contains all specified cities.

Country in which Gregory traveled consists of n cities and m undirected roads between them. For each road Ostap knows the time it takes to travel it, and the "shortest" word above is with respect to those times.

It is guaranteed that there exists some shortest path going through all specified cities.

Input
First line contains two integers n, m (1 ≤ n, m ≤ 105). Each of the m following lines contains a description of a single road ai, bi, ti (ai ≠ bi, 1 ≤ ai, bi ≤ n, 1 ≤ ti ≤ 104) means Gregory can go between ai and bi by road and that will take ti seconds. The next line contains k — the number of cities that Ostap knows Gregory has visited. The last line contains a list of these cities. All cities in that list are distinct.

Output
On the first line output the number of roads in the sought shortest path. On the second line output the list of road numbers (numbered in the order of appearing in the input) in the order of that shortest path. If there are many solutions, output any.

Example(s)
sample input
sample output
6 6
1 2 2
2 6 2
1 3 1
3 4 1
4 5 1
5 6 1
3
5 1 3
3
3 4 5

sample input
sample output
6 6
1 2 2
2 6 2
1 3 1
3 4 1
4 5 1
5 6 1
2
1 6
2
1 2

<|response|>
1. Abridged Problem Statement
Given an undirected graph of n vertices and m weighted edges (positive weights), and a list of k distinct "important" vertices. It is guaranteed that there exists some pair of vertices s,t such that one of the shortest paths from s to t passes through all k important vertices. Find any such shortest path and output the sequence of edge‐indices on it.

2. Key Observations
- Any subpath of a shortest path is itself a shortest path.
- Among all pairs of important vertices, let A and B be a farthest pair in terms of shortest‐path distance in the full graph. Then any shortest path from A to B must pass through every important vertex. (Otherwise you could replace one endpoint by that missed vertex and get an even larger distance.)
- Once we fix A, Dijkstra from A gives us dist[v] = distance(A→v). The unique shortest‐path DAG (directed acyclic graph) from A is formed by orienting each edge (u–v, w) as u→v if dist[v] = dist[u] + w. Every path in this DAG is a shortest path from A to its endpoint.
- We now need to pick a path in this DAG from A to B that visits all important vertices. Mark each important vertex with weight 1, others weight 0, and do a DP on the DAG to compute for every u: dp[u] = maximum number of important vertices collectable on any DAG‐path from u to any sink. Since we know there is a path from A to B that visits all k, dp[A] = k.
- Finally, we reconstruct by walking from A to B, at each node u choosing an outgoing edge u→v for which dp[u] = isImportant[u] + dp[v].

3. Full Solution Approach
Step 1: Read graph, important list, zero‐index vertices.
Step 2: Run Dijkstra from the first important vertex to get distances dist0[]. Among the k important vertices, pick A = arg max dist0[].
Step 3: Run Dijkstra from A to get distA[]. Among important vertices, pick B = arg max distA[].
Step 4: Build the shortest‐path DAG from A: for each undirected edge (u,v,w) if distA[v] == distA[u] + w then add u→v; if distA[u] == distA[v] + w then add v→u.
Step 5: Mark an array visited[v]=true if v is important else false. Do a memoized DFS on the DAG:
 dp[u] = visited[u] + max_{u→v in DAG}(dp[v])
This runs in O(n+m). In particular, dp[A] will equal k.
Step 6: Reconstruct one path from A to B: set cur=A and while cur≠B choose any edge cur→v in the DAG with dp[cur] = visited[cur] + dp[v], record its original edge‐index, and set cur=v.
Step 7: Output the collected edge‐indices (converted to 1‐based).

Overall complexity: Two Dijkstra's O(m log n), plus O(m+n) for the DAG and DP.

4. C++ implementation with comments

```cpp
#include <bits/stdc++.h>

using namespace std;

template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second;
}

template<typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& x) {
    return in >> x.first >> x.second;
}

template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    for(auto& x: a) {
        in >> x;
    }
    return in;
};

template<typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
    for(auto x: a) {
        out << x << ' ';
    }
    return out;
};

int n, m, k;
vector<tuple<int, int, int>> edges;
vector<vector<pair<int, int>>> adj;
vector<int> important;

void read() {
    cin >> n >> m;
    edges.resize(m);
    adj.assign(n, {});
    for(int i = 0; i < m; i++) {
        int u, v, w;
        cin >> u >> v >> w;
        u--, v--;
        adj[u].push_back({v, i});
        adj[v].push_back({u, i});
        edges[i] = {u, v, w};
    }

    cin >> k;
    important.resize(k);
    cin >> important;
    for(int &v: important) {
        v--;
    }
}

vector<int> dijkstra(int src) {
    vector<int> dist(n, 1e9);
    dist[src] = 0;
    priority_queue<
        pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>>
        pq;
    pq.push({0, src});
    while(!pq.empty()) {
        auto [d, u] = pq.top();
        pq.pop();
        if(dist[u] < d) {
            continue;
        }
        for(auto [v, i]: adj[u]) {
            auto [_, __, w] = edges[i];
            if(dist[v] > dist[u] + w) {
                dist[v] = dist[u] + w;
                pq.push({dist[v], v});
            }
        }
    }
    return dist;
}

vector<vector<pair<int, int>>> build_shortest_path_dag(const vector<int>& dist
) {
    vector<vector<pair<int, int>>> dag(n);
    for(int u = 0; u < n; u++) {
        for(auto [v, i]: adj[u]) {
            auto [_, __, w] = edges[i];
            if(dist[v] == dist[u] + w) {
                dag[u].push_back({v, i});
            }
        }
    }
    return dag;
}

int get_furthest(const vector<int>& dist) {
    int endpoint = important[0];
    for(int v: important) {
        if(dist[v] > dist[endpoint]) {
            endpoint = v;
        }
    }
    return endpoint;
}

void solve() {
    // Find a single shortest path that passes through every important city.
    // Any such path, restricted to the important cities, is itself a
    // shortest path between its two extreme important endpoints, so we first
    // locate those endpoints with a double-sweep (like a tree diameter):
    // Dijkstra from any important city gives the farthest important city
    // (endpoint); Dijkstra from endpoint gives the other extreme.
    //
    // On the shortest-path DAG rooted at endpoint (edges (u, v) with
    // dist[v] == dist[u] + w), dp[u] = the maximum number of important
    // cities on a shortest path from endpoint to u. We then walk from
    // endpoint to other_endpoint always stepping to a successor that
    // preserves the optimal dp value, which yields a shortest path
    // collecting all important cities, and we record the edge ids taken.

    vector<int> dist = dijkstra(important[0]);
    int endpoint = get_furthest(dist);

    dist = dijkstra(endpoint);
    vector<int> dp(n, -1);
    vector<bool> visited(n, false);
    for(int v: important) {
        visited[v] = true;
    }

    vector<vector<pair<int, int>>> dag = build_shortest_path_dag(dist);
    function<int(int)> dfs = [&](int u) {
        if(dp[u] != -1) {
            return dp[u];
        }
        dp[u] = visited[u];
        for(auto [v, i]: dag[u]) {
            dp[u] = max(dp[u], dfs(v) + visited[u]);
        }
        return dp[u];
    };

    int other_endpoint = get_furthest(dist);
    int start = endpoint;

    vector<int> ans;
    while(start != other_endpoint) {
        for(auto [v, i]: dag[start]) {
            if(dfs(start) == dfs(v) + visited[start]) {
                ans.push_back(i + 1);
                start = v;
                break;
            }
        }
    }

    cout << ans.size() << '\n';
    cout << ans << '\n';
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

    int T = 1;
    // cin >> T;
    for(int test = 1; test <= T; test++) {
        read();
        // cout << "Case #" << test << ": ";
        solve();
    }

    return 0;
}
```

5. Python implementation with detailed comments

```python
import sys
import threading
import heapq

def main():
    sys.setrecursionlimit(10**7)
    input = sys.stdin.readline

    n, m = map(int, input().split())
    # Store edges as (u, v, w). 0-based.
    edges = []
    # adjacency: for each u, list of (v, edge_index)
    adj = [[] for _ in range(n)]
    for i in range(m):
        u, v, w = map(int, input().split())
        u -= 1; v -= 1
        edges.append((u, v, w))
        adj[u].append((v, i))
        adj[v].append((u, i))

    k = int(input())
    important = list(map(lambda x: int(x)-1, input().split()))
    is_imp = [0]*n
    for v in important:
        is_imp[v] = 1

    # Dijkstra from src -> returns dist array
    def dijkstra(src):
        INF = 10**18
        dist = [INF]*n
        dist[src] = 0
        pq = [(0, src)]
        while pq:
            d,u = heapq.heappop(pq)
            if d > dist[u]:
                continue
            for v, ei in adj[u]:
                w = edges[ei][2]
                nd = d + w
                if nd < dist[v]:
                    dist[v] = nd
                    heapq.heappush(pq, (nd, v))
        return dist

    # Find the important node farthest from src
    def get_farthest(dist):
        best = important[0]
        for v in important:
            if dist[v] > dist[best]:
                best = v
        return best

    # 1) pick initial, find A
    dist0 = dijkstra(important[0])
    A = get_farthest(dist0)
    # 2) from A, find B
    distA = dijkstra(A)
    B = get_farthest(distA)

    # 3) build shortest-path DAG wrt distA
    dag = [[] for _ in range(n)]
    for u in range(n):
        for v, ei in adj[u]:
            w = edges[ei][2]
            if distA[v] == distA[u] + w:
                dag[u].append((v, ei))

    # 4) DP[u] = max important-nodes collectable from u
    dp = [-1]*n
    def dfs(u):
        if dp[u] != -1:
            return dp[u]
        best = is_imp[u]
        for v, ei in dag[u]:
            best = max(best, is_imp[u] + dfs(v))
        dp[u] = best
        return best

    dfs(A)

    # 5) Reconstruct path from A to B
    ans = []
    cur = A
    while cur != B:
        for v, ei in dag[cur]:
            if dp[cur] == is_imp[cur] + dp[v]:
                ans.append(ei+1)  # convert to 1-based
                cur = v
                break

    # Output
    print(len(ans))
    print(*ans)

if __name__ == "__main__":
    main()
```

Explanation of key steps is in the comments. This runs in O(m log n + n + m) time and O(n+m) memory, which is suitable for n,m up to 1e5.
