## 1. Abridged Problem Statement

There are `N` towns, each with 4 gates, so there are `4N` gates total. Every gate is connected by exactly one bidirectional road to a gate of another town, giving `2N` roads.

A courier starts in the capital town `1`, may move freely between gates inside a town, but outside towns may only travel along roads. The courier must pass every gate exactly once and return to the capital.

Determine whether such a route exists. If yes, output `"Yes"` and any valid ordering of all `4N` gates in the order the courier passes them; otherwise output `"No"`.

---

## 2. Detailed Editorial

### Key observation

Each gate has exactly one outside road. Inside a town, the courier can move between any two gates freely.

Suppose we decide, for each town, how the courier moves inside that town. Since each town has 4 gates, we can pair its gates into two internal transitions. For example, in a town with gates:

```text
a b c d
```

we may choose internal pairs:

```text
(a, b) and (c, d)
```

Then every gate has exactly two incident “route edges”:

1. its fixed outside road,
2. one chosen internal connection inside its town.

So if we combine:

- all given road edges,
- our chosen internal pairings,

we get a graph where every gate has degree `2`.

A finite graph where every vertex has degree `2` is a disjoint union of cycles.

Therefore, if we can choose the internal pairings so that all gates belong to one single cycle, that cycle is a valid courier route.

---

### Initial construction

For every town, initially pair its four gates as:

```text
(4i - 3, 4i - 2)
(4i - 1, 4i)
```

Together with the roads, this creates several disjoint cycles.

If there is only one cycle, we are done.

Otherwise, we try to merge cycles.

---

### Merging two cycles

Focus on the cycle containing gate `1`, i.e. a cycle passing through the capital.

Consider a town. Its four gates are split into two internal pairs.

Because internal pairs are edges in our current graph, each pair lies entirely in some cycle.

So, with respect to the capital cycle, a town can have:

- `0` gates in the capital cycle,
- `2` gates in the capital cycle,
- `4` gates in the capital cycle.

If some town has exactly `2` gates in the capital cycle, then:

- one internal pair belongs to the capital cycle,
- the other internal pair belongs to some other cycle.

Let those gates be:

```text
capital cycle gates: x1, x2
other cycle gates:   y1, y2
```

Currently the internal pairings are:

```text
x1 -- x2
y1 -- y2
```

We replace them with:

```text
x1 -- y1
x2 -- y2
```

This operation joins the two cycles into one larger cycle.

So the number of cycles decreases by one.

---

### What if no such town exists?

If more than one cycle exists, but no town has exactly two gates in the capital cycle, then every town has either all 4 gates or no gates in the capital cycle.

That means there is no road connecting a town in the capital component to a town outside it; otherwise the road endpoint outside would also belong to the capital cycle.

So the towns are disconnected from the capital, and the courier cannot visit all gates.

Therefore the answer is `"No"`.

---

### Algorithm

1. Read all roads.
2. Initialize internal gate pairings town by town:
   - `(1,2), (3,4)`
   - `(5,6), (7,8)`
   - etc.
3. Repeatedly:
   - Label all current cycles.
   - If there is only one cycle, output it.
   - Otherwise, find a town with exactly two gates in the cycle containing gate `1`.
   - If none exists, output `"No"`.
   - Otherwise, re-pair the four gates of that town to merge the capital cycle with another cycle.
4. After one cycle remains, traverse it and output the gate order.

---

### Complexity

There are `4N` gates.

Each cycle-labeling pass is `O(N)`.  
Each merge reduces the number of cycles by one, and there are at most `O(N)` cycles.

So the total complexity is:

```text
O(N^2)
```

With `N ≤ 1000`, this is easily fast enough.

---

## 3. Provided C++ Solution with Detailed Comments

```cpp
// Include almost all standard C++ library headers.
#include <bits/stdc++.h>

// Use the standard namespace to avoid writing std:: everywhere.
using namespace std;

// Overload output operator for pairs.
// This allows writing cout << pair<int,int>{a,b}.
template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    // Print the two elements separated by a space.
    return out << x.first << ' ' << x.second;
}

// Overload input operator for pairs.
// This allows writing cin >> pair_variable.
template<typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& x) {
    // Read first and second elements.
    return in >> x.first >> x.second;
}

// Overload input operator for vectors.
// This reads all existing elements of a vector.
template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    // Iterate over all elements by reference and read them.
    for(auto& x: a) {
        in >> x;
    }

    // Return the stream to allow chaining.
    return in;
};

// Overload output operator for vectors.
// This prints all elements separated by spaces.
template<typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
    // Iterate over elements by value.
    for(auto x: a) {
        // Print current element followed by a space.
        out << x << ' ';
    }

    // Return the stream to allow chaining.
    return out;
};

// Number of towns.
int n;

// road[g] is the gate connected to gate g by the fixed outside road.
vector<int> road;

// Reads the input.
void read() {
    // Read number of towns.
    cin >> n;

    // There are 4*n gates, numbered from 1 to 4*n.
    // Index 0 is unused.
    road.assign(4 * n + 1, 0);

    // There are exactly 2*n roads because every one of 4*n gates
    // has degree 1 in the road matching.
    for(int i = 0; i < 2 * n; i++) {
        // Read the two gates connected by this road.
        int a, b;
        cin >> a >> b;

        // Store the bidirectional road.
        road[a] = b;
        road[b] = a;
    }
}

// Solves the problem.
void solve() {
    // Gates are vertices.
    // road[g] is the fixed outside-road partner of gate g.
    //
    // We will additionally choose one internal partner mate[g] inside the same town.
    // Then every gate has degree exactly 2:
    //   - one road edge,
    //   - one internal mate edge.
    //
    // Therefore the graph becomes a disjoint union of cycles.
    // We want to choose mate[] so that there is exactly one cycle containing all gates.

    // Total number of gates.
    int g = 4 * n;

    // mate[x] will be the internal gate paired with x in the same town.
    vector<int> mate(g + 1);

    // Start with an arbitrary internal pairing for every town.
    for(int town = 0; town < n; town++) {
        // Gates of this town are:
        // base, base+1, base+2, base+3.
        int base = 4 * town + 1;

        // Pair first two gates.
        mate[base] = base + 1;
        mate[base + 1] = base;

        // Pair last two gates.
        mate[base + 2] = base + 3;
        mate[base + 3] = base + 2;
    }

    // comp[x] will store the cycle/component index of gate x
    // in the current degree-2 graph formed by road[] and mate[].
    vector<int> comp(g + 1);

    // Lambda function that labels all current cycles.
    auto label = [&]() {
        // Mark all gates as unvisited.
        fill(comp.begin(), comp.end(), -1);

        // Number of cycles found so far.
        int c = 0;

        // Try every gate as a possible start of an unvisited cycle.
        for(int s = 1; s <= g; s++) {
            // If already labeled, skip it.
            if(comp[s] != -1) {
                continue;
            }

            // Start traversing this cycle.
            int cur = s;

            // Previous gate in traversal.
            // -1 means there is no previous gate yet.
            int prev = -1;

            // Traverse until we return to starting gate s.
            do {
                // Mark current gate as belonging to component/cycle c.
                comp[cur] = c;

                // In the degree-2 graph, current gate has two neighbors:
                // road[cur] and mate[cur].
                //
                // To continue walking along the cycle, choose the neighbor
                // that is not the vertex we just came from.
                int nxt = (road[cur] != prev) ? road[cur] : mate[cur];

                // Move one step forward.
                prev = cur;
                cur = nxt;

            } while(cur != s);

            // Finished one cycle.
            c++;
        }

        // Return the number of cycles.
        return c;
    };

    // Keep merging cycles until there is exactly one cycle.
    while(label() != 1) {
        // Component index of the cycle containing gate 1.
        // This is the "capital cycle".
        int cap = comp[1];

        // base will store the first gate number of a useful town.
        // A useful town has exactly two gates in the capital cycle.
        int base = -1;

        // Search all towns.
        for(int town = 0; town < n && base == -1; town++) {
            // First gate of this town.
            int b = 4 * town + 1;

            // Count how many of this town's four gates are in the capital cycle.
            int cnt = 0;

            // Check four gates of the town.
            for(int k = 0; k < 4; k++) {
                cnt += (comp[b + k] == cap);
            }

            // If exactly two gates are in the capital cycle,
            // this town can be used to merge the capital cycle with another one.
            if(cnt == 2) {
                base = b;
            }
        }

        // If no such town exists, then the capital cycle is separated
        // from remaining cycles at the town level.
        // Therefore no route visiting every gate is possible.
        if(base == -1) {
            cout << "No\n";
            return;
        }

        // in_gate[] will contain the two gates of this town that are
        // currently in the capital cycle.
        int in_gate[2];

        // out_gate[] will contain the other two gates of this town,
        // currently belonging to some other cycle.
        int out_gate[2];

        // Counters for the two arrays.
        int ii = 0, oo = 0;

        // Split the four gates of the selected town into capital-cycle gates
        // and non-capital-cycle gates.
        for(int k = 0; k < 4; k++) {
            // Current gate.
            int gg = base + k;

            // Gate belongs to capital cycle.
            if(comp[gg] == cap) {
                in_gate[ii++] = gg;
            }
            // Gate belongs to another cycle.
            else {
                out_gate[oo++] = gg;
            }
        }

        // Re-pair the gates across the two cycles.
        //
        // Before:
        //   in_gate[0] -- in_gate[1] were internally paired,
        //   out_gate[0] -- out_gate[1] were internally paired.
        //
        // After:
        //   in_gate[0] -- out_gate[0],
        //   in_gate[1] -- out_gate[1].
        //
        // This splices two disjoint cycles into one larger cycle.
        mate[in_gate[0]] = out_gate[0];
        mate[out_gate[0]] = in_gate[0];

        mate[in_gate[1]] = out_gate[1];
        mate[out_gate[1]] = in_gate[1];
    }

    // If we reach this point, all gates are in one cycle.
    // Now traverse that cycle and output its gate order.

    // Stores the final gate order.
    vector<int> order;

    // Reserve memory for efficiency.
    order.reserve(g);

    // Start traversal from gate 1.
    int cur = 1;

    // There is no previous gate initially.
    int prev = -1;

    // Traverse the unique cycle until we return to gate 1.
    do {
        // Add current gate to answer.
        order.push_back(cur);

        // Choose the next neighbor in the cycle.
        // Avoid going immediately back to the previous gate.
        int nxt = (road[cur] != prev) ? road[cur] : mate[cur];

        // Advance.
        prev = cur;
        cur = nxt;

    } while(cur != 1);

    // Output success.
    cout << "Yes\n";

    // Output the route.
    cout << order << "\n";
}

// Main function.
int main() {
    // Speed up C++ input/output.
    ios_base::sync_with_stdio(false);

    // Untie cin from cout for faster input.
    cin.tie(nullptr);

    // The problem has one test case.
    int T = 1;

    // Multiple test cases are not used here.
    // cin >> T;

    // Solve each test case.
    for(int test = 1; test <= T; test++) {
        // Read input.
        read();

        // Solve and print answer.
        solve();
    }

    // End program successfully.
    return 0;
}
```

---

## 4. Python Solution with Detailed Comments

```python
import sys


def solve():
    # Read all integers from standard input.
    data = list(map(int, sys.stdin.buffer.read().split()))

    # If input is empty, do nothing.
    if not data:
        return

    # First number is the number of towns.
    n = data[0]

    # Total number of gates.
    g = 4 * n

    # road[x] will be the gate connected to x by the fixed outside road.
    # Index 0 is unused because gates are numbered from 1.
    road = [0] * (g + 1)

    # The next 2*n pairs describe the roads.
    idx = 1
    for _ in range(2 * n):
        # Read road endpoints.
        a = data[idx]
        b = data[idx + 1]
        idx += 2

        # Store bidirectional road.
        road[a] = b
        road[b] = a

    # mate[x] is the chosen internal partner of gate x inside the same town.
    mate = [0] * (g + 1)

    # Initially pair gates in every town as:
    # (base, base+1) and (base+2, base+3).
    for town in range(n):
        # First gate of this town.
        base = 4 * town + 1

        # First internal pair.
        mate[base] = base + 1
        mate[base + 1] = base

        # Second internal pair.
        mate[base + 2] = base + 3
        mate[base + 3] = base + 2

    # comp[x] stores which current cycle gate x belongs to.
    comp = [-1] * (g + 1)

    def label_cycles():
        """
        Label all cycles in the current graph formed by:
        - fixed road edges road[x],
        - chosen internal edges mate[x].

        Since every gate has degree exactly 2, the graph is a union of cycles.

        Returns:
            number of cycles
        """

        # Reset all component labels.
        for i in range(g + 1):
            comp[i] = -1

        # Number of cycles found.
        cycle_count = 0

        # Try every gate as a starting point.
        for start in range(1, g + 1):
            # Skip already labeled gates.
            if comp[start] != -1:
                continue

            # Traverse this cycle.
            cur = start

            # Previous vertex in traversal.
            # 0 means "none"; no real gate has number 0.
            prev = 0

            while True:
                # Mark current gate with current cycle id.
                comp[cur] = cycle_count

                # The two possible neighbors are road[cur] and mate[cur].
                # Continue to the neighbor that is not the previous vertex.
                if road[cur] != prev:
                    nxt = road[cur]
                else:
                    nxt = mate[cur]

                # Advance along the cycle.
                prev, cur = cur, nxt

                # Once we return to start, this cycle is complete.
                if cur == start:
                    break

            # One full cycle has been labeled.
            cycle_count += 1

        return cycle_count

    # Repeatedly merge cycles until only one remains.
    while True:
        # Label current cycles.
        cycles = label_cycles()

        # If there is exactly one cycle, we have a valid route.
        if cycles == 1:
            break

        # Cycle containing gate 1, which belongs to the capital.
        capital_cycle = comp[1]

        # We need a town with exactly two gates in the capital cycle.
        chosen_base = -1

        # Search all towns.
        for town in range(n):
            # First gate of this town.
            base = 4 * town + 1

            # Count how many gates of this town lie in the capital cycle.
            cnt = 0
            for k in range(4):
                if comp[base + k] == capital_cycle:
                    cnt += 1

            # Such a town lets us splice the capital cycle with another cycle.
            if cnt == 2:
                chosen_base = base
                break

        # If no such town exists, the remaining cycles are disconnected
        # from the capital part at the town level.
        if chosen_base == -1:
            print("No")
            return

        # Gates of chosen town that are in the capital cycle.
        inside = []

        # Gates of chosen town that are outside the capital cycle.
        outside = []

        # Split the four gates.
        for k in range(4):
            gate = chosen_base + k
            if comp[gate] == capital_cycle:
                inside.append(gate)
            else:
                outside.append(gate)

        # Re-pair the internal edges to merge two cycles.
        #
        # Old situation:
        #   inside[0] -- inside[1]
        #   outside[0] -- outside[1]
        #
        # New situation:
        #   inside[0] -- outside[0]
        #   inside[1] -- outside[1]
        mate[inside[0]] = outside[0]
        mate[outside[0]] = inside[0]

        mate[inside[1]] = outside[1]
        mate[outside[1]] = inside[1]

    # Now all gates form one single cycle.
    # Traverse it to produce the route.

    order = []

    # Start from gate 1.
    cur = 1

    # Previous gate; 0 means no previous gate.
    prev = 0

    while True:
        # Add current gate to output order.
        order.append(cur)

        # Continue along the cycle, avoiding the edge we just used.
        if road[cur] != prev:
            nxt = road[cur]
        else:
            nxt = mate[cur]

        # Move forward.
        prev, cur = cur, nxt

        # Stop when the cycle returns to gate 1.
        if cur == 1:
            break

    # Print answer.
    print("Yes")
    print(*order)


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

---

## 5. Compressed Editorial

Treat every gate as a vertex. Every gate already has one fixed road edge. Choose, inside each town, a perfect matching of its four gates; these are the courier’s internal moves. Then every gate has degree `2`, so the resulting graph is a disjoint union of cycles. We need one cycle containing all gates.

Start with arbitrary pairings inside each town: `(1,2),(3,4)` relative to that town. Label all cycles. If there is one cycle, output it.

Otherwise, look at the cycle containing gate `1`. If a town has exactly two gates in this cycle, its other two gates belong to another cycle. Replacing the two internal pairs by cross-pairs merges these two cycles. Repeat.

If no town has exactly two gates in the capital cycle while multiple cycles remain, then every town is either fully inside or fully outside the capital component, so there is no road from the capital part to the rest. Hence no complete route exists.

After all cycles are merged, traverse the unique degree-2 cycle and output the visited gates.

Complexity: `O(N^2)`.