<|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.

518. Kidnapping
Time limit per test: 0.5 second(s)
Memory limit: 262144 kilobytes
input: standard
output: standard

Berland's Police has a serious problem. A foreign ambassador arrived to Berland with an important mission, and his daughter was kidnapped just from the Royal Palace! Inspired by adventures of Erast Fandorin, the Police Chief developed the following ingenious plan.

The ambassador agrees to pay ransom, but only if the kidnappers allow his servant to visit the girl and ensure that she is alive. The kidnappers take the blindfolded servant into a coach and transport him to the secret place, where they keep the ambassador's daughter. Certainly, the role of the servant is certainly played by a secret agent of the Police. The Police Chief knows that when the coach is moving, the wheels are creaking once on each full rotation. So, by counting the number of creaks and multiplying it by the length of the rim, one can easily calculate the distance covered by the coach.

In spite of this brilliant idea, the affair turned to be much more difficult than it could be in a detective story. There are n intersections in the city numbered from 1 to n, some pairs of intersections are connected by bidirectional roads. The kidnappers agreed to take the "servant" to the secret place, and the servant is quite sure that this place is located at one of the intersections. Also the agent has calculated the lengths of roads between each pair of consecutive intersections on the route passed by the coach. But during the trip the agent was concentrated on counting creaks, so he could not remember in which directions the coach turned at the intersections.

Now the route probably couldn't be restored uniquely! Moreover, the agent has a suspicion that the kidnappers could intentionally pass the same intersection or even the same road more than once to confuse the Police.

Your task is to determine all possible locations of the secret place, given that the trip starts at the intersection number 1.

Input
The first line of the input contains a single integer n (2 ≤ n ≤ 200). Each of the next n lines contains n integers each. The i-th number in the j-th line lij is the length of the road between the i-th and the j-th intersections. If lij = 0 then the road doesn't exist.

It is guaranteed that 0 ≤ lij ≤ 200, lii = 0 and lij = lji. The next line contains one integer k (1 ≤ k ≤ 200) — the number of roads passed by the couch. The following line contains k integers r1, r2,..., rk (1 ≤ ri ≤ 200) — the lengths of roads between each pair of consecutive intersections on the route passed by the coach from the starting point to the secret place.

Output
To the first line of the output write m — the number of all possible locations of the secret place. The second line should contain the numbers of intersections in increasing order separated by spaces.

If there are no possible locations of the secret place, the output must contain the only integer 0.

Example(s)
sample input
sample output
4
0 1 2 0
1 0 1 0
2 1 0 2
0 0 2 0
3
1 1 2
3
1 3 4

<|response|>
1. Abridged problem statement
We have an undirected graph of n intersections (nodes 1..n) where each pair (i,j) may have a road of positive integer length lij (lij = 0 means no road). A coach starts at node 1 and traverses exactly k roads, producing a recorded sequence of lengths r₁…rₖ in order. The coach may revisit nodes or roads. We need to find all end-nodes where the coach could finish so that the sequence of edge lengths along the path exactly equals r₁…rₖ. Output the count of such nodes and their labels in increasing order; if none, output 0.

2. Key observations
- We only care about matching the given sequence of edge lengths, not the actual directions or nodes visited in between, beyond matching lengths.
- Revisiting nodes or edges is allowed, so we must consider all walks, not just simple paths.
- A naive enumeration of all walks of length k is exponential. Instead we use a DP/BFS over a state space of size O(n × (k+1)).
- Define state (u, p) meaning "the coach is at node u after matching p lengths of the sequence." From (u, p) with p<k, we can go to (v, p+1) for any neighbor v where the edge length between u and v equals rₚ₊₁.
- Mark each state visited to avoid reprocessing. When we reach any state (u, k), u is a valid end-node.

3. Full solution approach
- Read n and the n×n adjacency matrix graph[i][j].
- Read k and the sequence r[0…k−1].
- Create a 2D boolean array visited[n][k+1], all false initially.
- Use a queue of pairs (node, matched), initialize with (0, 0) (0-indexed) and mark visited[0][0] = true.
- While the queue is not empty:
  - Dequeue (u, p).
  - If p == k, record u in a result set and continue to next state.
  - Otherwise, look at every v from 0 to n-1. If graph[u][v] == r[p] and visited[v][p+1] is false, mark it true and enqueue (v, p+1).
- After BFS finishes, collect all recorded end-nodes (convert to 1-based), sort them, and print the count + list (or 0 if empty).
- Time complexity: O(n² × k), which for n,k ≤ 200 is about 8 · 10⁶ edge checks, acceptable in 0.5 s.

4. C++ implementation with detailed 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, k;
vector<vector<int>> graph;
vector<int> path;

void read() {
    cin >> n;
    graph.assign(n, vector<int>(n));
    cin >> graph;
    cin >> k;
    path.resize(k);
    cin >> path;
}

void solve() {
    // BFS over states (node, pos): we are at intersection node having
    // matched the first pos road lengths of the route. From (node, pos) we
    // step to any next_node whose connecting road length graph[node][next]
    // equals path[pos], advancing to (next, pos + 1). visited[node][pos]
    // prevents revisiting a state. Every state reaching pos == k marks node
    // as a possible secret place. We collect those nodes (1-indexed),
    // sorted ascending.

    deque<pair<int, int>> queue = {{0, 0}};
    vector<vector<bool>> visited(n, vector<bool>(k + 1, false));
    visited[0][0] = true;
    set<int> possible;

    while(!queue.empty()) {
        auto [node, pos] = queue.front();
        queue.pop_front();

        if(pos == k) {
            possible.insert(node);
            continue;
        }

        for(int next_node = 0; next_node < n; next_node++) {
            if(graph[node][next_node] == path[pos] &&
               !visited[next_node][pos + 1]) {
                visited[next_node][pos + 1] = true;
                queue.push_back({next_node, pos + 1});
            }
        }
    }

    cout << possible.size() << '\n';
    if(!possible.empty()) {
        bool first = true;
        for(int x: possible) {
            cout << (first ? "" : " ") << x + 1;
            first = false;
        }
        cout << '\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
from collections import deque

def solve():
    # Number of intersections
    n = int(input())
    # Read adjacency matrix: graph[i][j] = length between i+1 and j+1, or 0 if no road
    graph = [list(map(int, input().split())) for _ in range(n)]

    # Number of roads traveled
    k = int(input())
    # The exact sequence of road lengths observed
    path = list(map(int, input().split()))

    # BFS queue of states: (current_node, how_many_roads_matched)
    queue = deque([(0, 0)])  # start at node 0, matched 0 roads
    # visited[node][pos] marks we've reached `node` after matching pos edges
    visited = [[False] * (k + 1) for _ in range(n)]
    visited[0][0] = True

    possible = set()  # all end-nodes once pos == k

    while queue:
        node, pos = queue.popleft()

        # If we've matched the full sequence, record this node
        if pos == k:
            possible.add(node)
            continue

        # Try every neighbor for the next step
        needed_length = path[pos]
        for nxt in range(n):
            # If there's a road of the needed length and we haven't visited (nxt,pos+1)
            if graph[node][nxt] == needed_length and not visited[nxt][pos + 1]:
                visited[nxt][pos + 1] = True
                queue.append((nxt, pos + 1))

    # Prepare output
    possible = sorted(possible)
    if not possible:
        print(0)
    else:
        # Number of possible secret places
        print(len(possible))
        # Convert to 1-based indexing
        print(*[x + 1 for x in possible])

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