1. Abridged Problem Statement
Given N domino pieces, each marked with two numbers (0–6), arrange all pieces in a single line so that adjacent ends match. You may flip (rotate) any piece, swapping its two numbers. If such an arrangement exists, output a valid ordering: for each piece, print its original index (1-based) and “+” if kept as given or “-” if flipped. If no arrangement uses all pieces in one chain, print “No solution”.

2. Detailed Editorial

Overview
This is the classic problem of finding an Eulerian trail (or circuit) in a multigraph. Represent each domino side number (0–6) as a vertex, and each domino piece as an undirected edge between its two numbers. A valid domino chain that uses every piece exactly once corresponds to an Eulerian trail in this graph.

Key Observations
- A connected undirected graph has an Eulerian trail that uses every edge exactly once if and only if it has either 0 or 2 vertices of odd degree.
- If there are 0 odd-degree vertices, there exists an Eulerian circuit (starts and ends at the same vertex).
- If there are exactly 2 odd-degree vertices, there is an Eulerian path starting at one odd vertex and ending at the other.
- If more than 2 vertices have odd degree, no solution exists.

Algorithm Steps
1. Build the graph: 7 vertices (0–6), N undirected edges for the dominoes; track degrees.
2. Collect all vertices of odd degree.
3. To unify the treatment of trail vs. circuit, pair up odd vertices arbitrarily and add “fake” edges between each pair. This makes all vertex degrees even, so the augmented graph has an Eulerian circuit that can be found by Hierholzer’s algorithm.
4. Run Hierholzer’s algorithm to decompose the augmented graph into cycles:
   - Maintain adjacency lists of “half-edges” (directed representations), each identified by a unique integer.
   - Traverse unused half-edges recursively, marking edges used and appending them to the current path.
5. Extract the single sequence of original domino edges from the found cycles by cutting at the fake edges; this yields one or more trails in the original graph.
6. If exactly one trail is produced, it covers all original dominoes; translate half-edge identifiers back to domino indices and orientations (“+” or “−”) and output them in order. Otherwise, print “No solution”.

Complexities
- N ≤ 100 edges, vertices = 7.
- Building the graph, pairing odd vertices, and Hierholzer’s algorithm all run in O(N).

3. C++ Solution
```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;
};

class EulerianPaths {
  private:
    int n, m;
    vector<vector<pair<int, int>>> adj;
    vector<pair<int, int>> edges;
    vector<int> deg;

    void dfs(int u, vector<int>& path, vector<bool>& used, vector<int>& po) {
        for(; po[u] < (int)adj[u].size();) {
            int idx = po[u]++;
            if(!used[adj[u][idx].second >> 1]) {
                used[adj[u][idx].second >> 1] = true;
                dfs(adj[u][idx].first, path, used, po);
                path.push_back(adj[u][idx].second);
            }
        }
    }

  public:
    EulerianPaths(int _n) : n(_n), m(0) {
        adj.assign(n + 1, {});
        deg.assign(n + 1, 0);
    }

    void add_edge(int u, int v) {
        adj[u].push_back({v, m * 2});
        adj[v].push_back({u, m * 2 + 1});
        edges.push_back({u, v});
        deg[u]++;
        deg[v]++;
        m++;
    }

    vector<vector<int>> find_paths() {
        vector<bool> used(m, false);
        vector<int> po(n + 1, 0);

        vector<int> odd_vertices;
        for(int i = 0; i <= n; i++) {
            if(deg[i] % 2 == 1) {
                odd_vertices.push_back(i);
            }
        }

        int total_edges = m;
        for(int i = 0; i < (int)odd_vertices.size() / 2; i++) {
            int u = odd_vertices[2 * i], v = odd_vertices[2 * i + 1];
            adj[u].push_back({v, 2 * total_edges});
            adj[v].push_back({u, 2 * total_edges + 1});
            total_edges++;
            used.push_back(false);
            edges.push_back({u, v});
        }

        vector<vector<int>> paths;
        for(int u = 0; u <= n; u++) {
            if(!adj[u].empty()) {
                vector<int> path;
                dfs(u, path, used, po);
                if(!path.empty()) {
                    // Rotate the path so that we always start with a fake edge
                    // if there is at least one.
                    auto it = find_if(path.begin(), path.end(), [&](int x) {
                        return x >= 2 * m;
                    });
                    if(it != path.end()) {
                        rotate(path.begin(), it, path.end());
                    }

                    vector<int> current_path;
                    for(int x: path) {
                        if(x < 2 * m) {
                            current_path.push_back(x);
                        } else if(!current_path.empty()) {
                            paths.push_back(current_path);
                            current_path.clear();
                        }
                    }
                    if(!current_path.empty()) {
                        paths.push_back(current_path);
                    }
                }
            }
        }

        return paths;
    }

    pair<int, int> get_edge(int edge_i) {
        if(edge_i & 1) {
            return edges[edge_i >> 1];
        } else {
            return {edges[edge_i >> 1].second, edges[edge_i >> 1].first};
        }
    }

    vector<pair<int, int>> get_path_edges(const vector<int>& path) {
        vector<pair<int, int>> result;
        for(int edge_i: path) {
            result.push_back(get_edge(edge_i));
        }
        return result;
    }
};

int m;
vector<pair<int, int>> dominos;

void read() {
    cin >> m;
    dominos.resize(m);
    cin >> dominos;
}

void solve() {
    // - Model the domino set as a multigraph on vertices 0..6 (the pip
    //   values), where each domino (a, b) is an undirected edge a-b. Arranging
    //   all dominoes in a chain that matches touching halves is exactly an
    //   Eulerian path that uses every edge once.
    //
    // - EulerianPaths pairs up odd-degree vertices with extra fake edges so an
    //   Eulerian circuit always exists, runs Hierholzer's algorithm, then cuts
    //   the circuit at the fake edges to recover the real path segments. A
    //   single segment covering all dominoes means a valid line exists;
    //   otherwise there is no solution.
    //
    // - For each used edge we print its 1-based domino index and a sign: '+'
    //   if traversed in the stored orientation, '-' if reversed.

    EulerianPaths ep(6);
    for(int i = 0; i < m; i++) {
        ep.add_edge(dominos[i].first, dominos[i].second);
    }

    auto paths = ep.find_paths();
    if(paths.size() == 1) {
        for(int edge_i: paths[0]) {
            cout << (edge_i >> 1) + 1 << ' '
                 << (ep.get_edge(edge_i) == dominos[edge_i >> 1] ? '+' : '-')
                 << '\n';
        }
    } else {
        cout << "No solution\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;
}
```

4. Python Solution
```python
import sys
sys.setrecursionlimit(10**7)

def read_input():
    """Read N and the list of dominoes."""
    N = int(sys.stdin.readline().strip())
    dominos = [tuple(map(int, sys.stdin.readline().split())) for _ in range(N)]
    return N, dominos

class Eulerian:
    def __init__(self, max_vertex):
        self.n = max_vertex
        self.adj = [[] for _ in range(self.n+1)]
        self.edges = []     # list of (u,v)
        self.deg = [0]*(self.n+1)
        self.m = 0

    def add_edge(self, u, v):
        """Add an undirected edge as two half-edges with IDs 2*m, 2*m+1."""
        heid1 = 2*self.m
        heid2 = 2*self.m+1
        self.adj[u].append((v, heid1))
        self.adj[v].append((u, heid2))
        self.edges.append((u, v))
        self.deg[u] += 1
        self.deg[v] += 1
        self.m += 1

    def find_trails(self):
        """Pair odd vertices, run Hierholzer, and extract trails."""
        used = [False]*self.m
        # Find odd-degree vertices
        odd = [v for v in range(self.n+1) if self.deg[v] % 2 == 1]
        # If >2 odd vertices, no solution
        if len(odd) > 2:
            return None

        # Pair odd vertices (if 2) by adding a fake edge
        tot = self.m
        if len(odd) == 2:
            u, v = odd
            # Fake half-edges IDs: 2*tot, 2*tot+1
            self.adj[u].append((v, 2*tot))
            self.adj[v].append((u, 2*tot+1))
            self.edges.append((u, v))
            used.append(False)
            tot += 1

        # Prepare for Hierholzer
        ptr = [0]*(self.n+1)
        trails = []

        def dfs(u, path):
            """Recursively follow unused halves to build a cycle."""
            while ptr[u] < len(self.adj[u]):
                v, heid = self.adj[u][ptr[u]]
                ptr[u] += 1
                idx = heid >> 1
                if not used[idx]:
                    used[idx] = True
                    dfs(v, path)
                    path.append(heid)

        # Run DFS from every vertex that has edges
        for start in range(self.n+1):
            if ptr[start] < len(self.adj[start]):
                cycle = []
                dfs(start, cycle)
                if not cycle:
                    continue
                # If we added a fake edge, rotate so fake is first
                fake_id = 2*self.m
                for i, he in enumerate(cycle):
                    if he >= fake_id:
                        cycle = cycle[i:] + cycle[:i]
                        break
                # Split at fake edges (if any) to form actual trails
                cur = []
                for he in cycle:
                    if he < 2*self.m:
                        cur.append(he)
                    else:
                        if cur:
                            trails.append(cur)
                            cur = []
                if cur:
                    trails.append(cur)

        # We need exactly one trail using all original edges
        if len(trails) == 1 and len(trails[0]) == self.m:
            return trails[0]
        else:
            return None

    def oriented_edge(self, heid):
        """Given half-edge ID, return oriented (u,v)."""
        idx = heid >> 1
        u, v = self.edges[idx]
        # if heid is even, direction is u->v; if odd, v->u
        if heid & 1:
            return (v, u)
        else:
            return (u, v)


def main():
    N, dominos = read_input()
    E = Eulerian(6)
    for u, v in dominos:
        E.add_edge(u, v)

    trail = E.find_trails()
    if trail is None:
        print("No solution")
        return

    # Output each edge's index and orientation
    for heid in trail:
        idx = (heid >> 1) + 1          # 1-based domino index
        u,v = E.oriented_edge(heid)
        # Check if this matches the original orientation
        if (u, v) == dominos[heid >> 1]:
            print(idx, '+')
        else:
            print(idx, '-')

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

5. Compressed Editorial
- Model numbers 0–6 as vertices, dominoes as edges.
- An arrangement is an Eulerian trail/circuit in this multigraph.
- Only 0 or 2 vertices of odd degree allow a trail; otherwise, no solution.
- Pair odd vertices (if any) with a fake edge to make all degrees even.
- Use Hierholzer’s algorithm on the augmented graph to find a circuit.
- Split at fake edges to recover the single required trail of original dominoes.
- Output edge indices with “+” or “−” depending on orientation; if impossible, print “No solution”.
