1. Concise 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.  
   - If their count > 2, output “No solution.”  
3. (Optional) 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 (or with a stack), 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 fake edges; this yields one or more trails in the original graph.  
6. If exactly one trail uses all original edges, 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. Provided C++ Solution with Detailed Comments  

```cpp
#include <bits/stdc++.h>
using namespace std;

// Overload << and >> for convenience with pairs and vectors
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 to find Eulerian paths/circuits in an undirected multigraph
class EulerianPaths {
  private:
    int n;                                 // Number of vertices
    int m;                                 // Number of original edges
    vector<vector<pair<int,int>>> adj;     // adj[u] = list of (neighbor, half-edge id)
    vector<pair<int,int>> edges;           // Original edges: edges[i] = (u,v)
    vector<int> deg;                       // Degree of each vertex

    // Recursive DFS for Hierholzer's algorithm
    // u = current vertex
    // path = collects half-edge ids in reverse order
    // used = marks original edges (by index) as used
    // po = position pointers in each adj[u] to avoid revisiting
    void dfs(int u, vector<int>& path, vector<bool>& used, vector<int>& po) {
        // Explore all remaining half-edges from u
        while (po[u] < (int)adj[u].size()) {
            auto [v, heid] = adj[u][po[u]++];       // get next neighbor and half-edge id
            int eidx = heid >> 1;                  // original edge index = heid/2
            if (!used[eidx]) {
                used[eidx] = true;                 // mark this undirected edge used
                dfs(v, path, used, po);            // continue from v
                path.push_back(heid);              // record the traversed half-edge
            }
        }
    }

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

    // Add an undirected edge between u and v
    void add_edge(int u, int v) {
        // Each undirected edge becomes two half-edges: id = 2*m and 2*m+1
        adj[u].push_back({v, 2*m});
        adj[v].push_back({u, 2*m+1});
        edges.push_back({u, v});
        deg[u]++; deg[v]++;
        m++;
    }

    // Find all maximal trails after pairing odd vertices
    vector<vector<int>> find_paths() {
        vector<bool> used(m, false);            // track used original edges
        vector<int> po(n + 1, 0);               // current adjacency pointer for each vertex

        // 1) Identify odd-degree vertices
        vector<int> odd;
        for(int u = 0; u <= n; u++)
            if (deg[u] % 2 == 1)
                odd.push_back(u);

        // 2) Pair up odd vertices and add fake edges to make all degrees even
        int tot = m;
        for(int i = 0; i < (int)odd.size()/2; i++) {
            int u = odd[2*i], v = odd[2*i + 1];
            adj[u].push_back({v, 2*tot});
            adj[v].push_back({u, 2*tot + 1});
            edges.push_back({u, v});          // store fake edge
            used.push_back(false);            // extend used[] for fake edges
            tot++;
        }

        // 3) Run Hierholzer to extract cycles/trails
        vector<vector<int>> paths;
        for(int u = 0; u <= n; u++) {
            if (adj[u].empty()) continue;
            vector<int> cycle;
            dfs(u, cycle, used, po);
            if (cycle.empty()) continue;

            // 4) Rotate cycle so that a fake edge (if any) is first
            auto it = find_if(cycle.begin(), cycle.end(),
                              [&](int x){ return x >= 2*m; }); 
            if (it != cycle.end())
                rotate(cycle.begin(), it, cycle.end());

            // 5) Split at fake edges to form maximal original-edge trails
            vector<int> curr;
            for(int heid : cycle) {
                if (heid < 2*m) {
                    curr.push_back(heid);
                } else {
                    if (!curr.empty()) {
                        paths.push_back(curr);
                        curr.clear();
                    }
                }
            }
            if (!curr.empty())
                paths.push_back(curr);
        }
        return paths;
    }

    // Given half-edge id, recover the oriented edge (u->v)
    pair<int,int> get_edge(int heid) {
        int idx = heid >> 1;                     // original edge index
        auto [u,v] = edges[idx];
        // if heid is even, direction is u->v; if odd, v->u
        return (heid & 1) ? make_pair(v,u) : make_pair(u,v);
    }
};

// Globals for input
int m;
vector<pair<int,int>> dominos;

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

// Solve one test case
void solve() {
    // Build EulerianPaths on vertices 0..6
    EulerianPaths ep(6);
    for(int i = 0; i < m; i++) {
        ep.add_edge(dominos[i].first, dominos[i].second);
    }

    // Extract trails
    auto paths = ep.find_paths();
    // We need exactly one trail that uses all m edges
    if (paths.size() == 1 && (int)paths[0].size() == m) {
        // Output each half-edge in the trail
        for(int heid : paths[0]) {
            int idx = (heid >> 1) + 1;                      // domino index (1-based)
            // Determine if orientation matches original or reversed
            auto e = ep.get_edge(heid);
            bool forward = (e == dominos[heid >> 1]);
            cout << idx << ' ' << (forward ? '+' : '-') << '\n';
        }
    } else {
        cout << "No solution\n";
    }
}

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

    read();
    solve();
    return 0;
}
```

4. Python Solution with Detailed Comments  

```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.”