1. Abridged Problem Statement  
Given an undirected graph with one capital (node 0) and n country towns arranged in a cycle (nodes 1…n). The capital is connected to every country town, and each town i is connected to towns i–1 and i+1 (modulo n). Count the number of walks of exactly m edges that start and end at the capital (node 0), allowing revisits.  
Input: integers n (3 ≤ n ≤ 1000) and m (0 ≤ m ≤ 5000).  
Output: the total number of such walks (exact value, no mod).  

2. Detailed Editorial  
We want the count of length-m walks from 0 back to 0 in this “wheel” graph. A direct DP on all n+1 states would be O(m·n), which in the worst case is about 5·10^6 transitions – borderline but possible in C++ with careful big-integer arithmetic. However, we can do much better in O(log m) big-integer operations by aggregating all the peripheral towns into one “supernode.”  

Define two macro-states:  
 • State 0 = at the capital.  
 • State 1 = at any country town (all towns merged).  

We track a 2-vector v_k = [ways to be at 0 after k steps, ways to be at super after k steps].  

Determine the 2×2 transition matrix A so that v_{k+1} = A · v_k:  
 - From state 0 you can only go to any of the n towns → super. That contributes n ways to go from state 0 to state 1: A[1][0] = n.  
 - From state 0 you cannot stay in 0: A[0][0] = 0.  
 - From state 1 (i.e. any town) you have exactly one edge back to 0 among the three edges of the cycle + spoke. But we compressed all towns, so that cluster contributes exactly 1 aggregated way: A[0][1] = 1.  
 - From state 1 you can also go to an adjacent town within the cycle (not to 0). Each town has two cycle‐neighbors. Aggregating again gives A[1][1] = 2.  

So  
    A = [[0, 1],  
         [n, 2]]  

with initial v₀ = [1, 0]. Then v_m = A^m · v₀, and the answer is v_m[0].  

We exponentiate the 2×2 matrix A in O(log m) multiprecisions big-integer multiplies and adds.  

3. Provided C++ Solution with Detailed Comments  
```cpp
#include <bits/stdc++.h>
#include <boost/multiprecision/cpp_int.hpp>
using namespace std;
// Use boost::multiprecision to handle arbitrarily large integers.
using BigInt = boost::multiprecision::cpp_int;

// Multiply two 2x2 matrices of BigInt.
array<array<BigInt,2>,2> mulMat(
    const array<array<BigInt,2>,2>& A,
    const array<array<BigInt,2>,2>& B)
{
    array<array<BigInt,2>,2> C;
    // C[i][j] = sum over k of A[i][k] * B[k][j]
    for(int i = 0; i < 2; ++i) {
        for(int j = 0; j < 2; ++j) {
            C[i][j] = 0;
            for(int k = 0; k < 2; ++k)
                C[i][j] += A[i][k] * B[k][j];
        }
    }
    return C;
}

// Fast exponentiation of 2x2 matrix to power p.
array<array<BigInt,2>,2> powMat(
    array<array<BigInt,2>,2> base,
    long long p)
{
    // Initialize result as identity matrix.
    array<array<BigInt,2>,2> result = { { {1,0}, {0,1} } };
    while(p > 0) {
        if (p & 1LL) {
            result = mulMat(result, base);
        }
        base = mulMat(base, base);
        p >>= 1;
    }
    return result;
}

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

    int n;
    long long m;
    cin >> n >> m;              // Read number of towns and length m.

    // Build the 2x2 transition matrix A.
    // A = [ [0, 1],
    //       [n, 2] ]
    array<array<BigInt,2>,2> A;
    A[0][0] = 0;     A[0][1] = 1;
    A[1][0] = n;     A[1][1] = 2;

    // Compute A^m by fast exponentiation.
    auto Am = powMat(A, m);

    // Initial state vector v0 = [1, 0]^T.
    // After m steps vm = Am * v0. We only need the first component:
    // vm[0] = Am[0][0]*1 + Am[0][1]*0 = Am[0][0].
    BigInt answer = Am[0][0];

    // Output the exact answer.
    cout << answer << "\n";
    return 0;
}
```

4. Python Solution with Detailed Comments  
```python
import sys

def matrix_mult_2x2(A, B):
    # Multiply two 2×2 matrices A and B.
    return [
        [
            A[0][0]*B[0][0] + A[0][1]*B[1][0],
            A[0][0]*B[0][1] + A[0][1]*B[1][1]
        ],
        [
            A[1][0]*B[0][0] + A[1][1]*B[1][0],
            A[1][0]*B[0][1] + A[1][1]*B[1][1]
        ]
    ]

def matrix_pow_2x2(A, power):
    # Fast exponentiation of 2×2 matrix A to integer power.
    # Start with identity matrix.
    result = [[1, 0], [0, 1]]
    while power > 0:
        if power & 1:
            result = matrix_mult_2x2(result, A)
        A = matrix_mult_2x2(A, A)
        power >>= 1
    return result

def solve(n, m):
    # Transition matrix A for the two macro-states:
    #    [ 0  1 ]
    #    [ n  2 ]
    A = [[0, 1], [n, 2]]
    # Raise A to the power m
    Am = matrix_pow_2x2(A, m)
    # Initial vector v0 = [1, 0]; after m steps vm = Am · v0
    # The number of ways to return to capital is vm[0].
    return Am[0][0]

def main():
    # Read n and m from stdin
    n, m = map(int, sys.stdin.readline().split())
    # Python int auto-scales to big integers
    print(solve(n, m))

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

5. Compressed Editorial  
- Merge all n peripheral towns into one “supernode.”  
- Model transitions between {capital, supernode} by a 2×2 matrix:  
      A = [[0, 1], [n, 2]].  
- Initial state v₀ = [1,0]. After m steps, vₘ = A^m · v₀; answer = vₘ[0].  
- Compute A^m in O(log m) using fast matrix exponentiation with big integers.