1. Abridged Problem Statement  
Given an integer n (3 ≤ n ≤ 5000), count the number of labeled connected undirected graphs on n vertices with exactly n edges (so each connected component has n vertices and n edges – equivalently, unicyclic connected graphs). Output the exact count without leading zeros.

2. Detailed Editorial  

Definition & Reformulation  
A connected graph on n vertices with exactly n edges is called a unicyclic graph: it has exactly one cycle, with trees (“branches”) possibly attached to the cycle vertices.

Counting by cycle‐size decomposition  
Let k be the length of the unique cycle (3 ≤ k ≤ n). We build any such graph in two independent steps:

 1. Choose which k vertices form the cycle and how they are cyclically connected.  
    - Number of ways to choose k vertices out of n: C(n,k).  
    - Number of distinct undirected cycles on k labeled vertices: (k−1)! / 2.  
    So total ways for the cycle itself: C(n,k) × ((k−1)!/2).

 2. Attach the remaining (n−k) vertices in trees hanging off the k-cycle vertices, so that the overall graph remains connected and no extra cycles appear.  
    - It is known (a generalization of Cayley’s formula) that the number of forests on n labeled vertices with exactly k specified roots (the cycle vertices) and each component containing exactly one root is k × n^(n−k−1).  
    Here those forests describe how the other n−k vertices connect (by unique paths) into roots on the cycle.

Multiplying and summing over k gives  
 Tₙ = ∑_{k=3}^n [ C(n,k)·((k−1)!/2) ] · [ k·n^(n−k−1) ]  
Simplify C(n,k)·(k−1)! = n!/(n−k)!/k, multiply by k/2 ⇒ n!/[2·(n−k)!]. Thus  
 Tₙ = (n!/2) · ∑_{k=3}^n n^(n−k−1) / (n−k)!  

Re‐index with j = n−k (so j = 0…n−3):  
 Tₙ = (n!/2) · ∑_{j=0}^{n−3} n^{j−1} / j!  

An O(n) iteration computes the inner sum and the prefactor (n!/2) with big‐integer arithmetic.

Implementation outline  
1. Compute f = (n−1)!/2 (the j=0 term).  
2. Accumulate ans = f.  
3. For j from 1 to n−3:  
     f = f × n / j  
     ans += f  
4. Print ans.

This runs in O(n) big‐integer multiplications/divisions, fast enough for n up to 5000.

3. C++ Solution with Detailed Comments  
```cpp
#include <bits/stdc++.h>
#include <boost/multiprecision/cpp_int.hpp>
using namespace std;
using big = boost::multiprecision::cpp_int;

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

    int n;
    cin >> n; 
    // We compute T_n = (n!/2) * sum_{j=0..n-3} n^{j-1} / j!,
    // but iteratively we build the j-th term of the sum in 'f':
    //   j=0: f = (n-1)! / 2
    //   then for each next j: f *= n; f /= j; 
    // and accumulate into 'ans'.

    // Compute f = (n-1)! / 2
    big f = 1;
    for (int i = 2; i < n; ++i) {
        f *= i;        // f = (n-1)!
    }
    f /= 2;            // now f = (n-1)! / 2

    big ans = f;       // initialize sum with j=0 term

    // Loop j from 1 to n-3 (inclusive),
    // each step updates f -> next term, then adds to ans.
    for (int j = 1; j <= n - 3; ++j) {
        f *= n;         // multiply numerator by n
        f /= j;         // divide denominator by j
        ans += f;       // add the new term
    }

    // Output the final big-integer result.
    cout << ans << "\n";
    return 0;
}
```

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

def main():
    n = int(sys.stdin.readline().strip())
    # We will compute:
    #   T_n = sum_{j=0 to n-3} f_j,
    # where f_0 = (n-1)! / 2, and
    #   f_{j} = f_{j-1} * n / j   for j >= 1

    # 1) Compute (n-1)! / 2
    f = 1
    for i in range(2, n):
        f *= i
    f //= 2  # now f = (n-1)! / 2

    ans = f  # start the sum with the j=0 term

    # 2) Iteratively build the remaining terms j=1..n-3
    for j in range(1, n - 2):  
        # Multiply by n and divide by j to move from f_{j-1} to f_j
        f = f * n // j
        ans += f

    # 3) Print the result
    print(ans)

if __name__ == "__main__":
    # Use a thread to raise recursion limits if needed
    threading.Thread(target=main).start()
```

5. Compressed Editorial  
Count unicyclic labeled graphs on n vertices by summing over cycle‐length k (3…n). For fixed k:  
  • choose k vertices and form a cycle in C(n,k)·((k−1)!/2) ways;  
  • attach the remaining n−k vertices as trees to the cycle in k·n^(n−k−1) ways.  
After algebraic simplification this yields  
  Tₙ = (n!/2)·∑_{j=0}^{n−3} n^{j−1}/j!.  
We compute the sum in O(n) big‐integer steps by initializing f₀=(n−1)!/2 and iterating fⱼ = f_{j−1}·n/j, accumulating into the final answer.