1. Abridged Problem Statement  
Given an undirected simple graph with N vertices and M edges, define its N×M incidence matrix A so that A[i][j]=1 if vertex i is one endpoint of edge j, and 0 otherwise. Compute the sum of all entries of the M×M matrix AᵀA.

2. Detailed Editorial  

Let A be the incidence matrix (size N×M). We need  
   S = sum of all entries of AᵀA.  

Observe that (AᵀA)[j][k] is the dot product of column j and column k of A. Column j has exactly two 1’s (the two endpoints of edge j), and similarly for column k. Thus:  
- When j = k, (AᵀA)[j][j] = 1·1 + 1·1 = 2 (the edge’s two endpoints).  
- When j ≠ k, (AᵀA)[j][k] = number of shared endpoints between edges j and k. Since the graph is simple, two distinct edges can share at most one vertex. If they share a vertex, the dot product is 1; otherwise it is 0.

Hence the total sum is  
  S = ∑_{j=1..M} (AᵀA)[j][j]  +  ∑_{j≠k} (AᵀA)[j][k]  
    = 2M  +  (number of ordered pairs of distinct edges that share a common endpoint).

To count the second term, look at each vertex v of degree d_v. Among the d_v edges incident to v, we can pick an ordered pair of distinct edges in d_v·(d_v−1) ways—and each such ordered pair contributes exactly 1 to ∑_{j≠k}(AᵀA)[j][k]. Summing over all vertices gives  
  ∑_{v=1..N} d_v·(d_v−1).  

Therefore  
  S = 2M + ∑_{v=1..N} d_v·(d_v−1).  

Implementation steps:  
1) Read N, M.  
2) Initialize an array deg[1..N] to zero.  
3) For each of the M edges (u, v): increment deg[u] and deg[v].  
4) Compute ans = 2*M + ∑ deg[i]*(deg[i]−1).  
5) Output ans.

Time complexity: O(N+M). Memory: O(N).

3. Provided C++ Solution with Detailed Comments  
```cpp
#include <bits/stdc++.h>
using namespace std;

// Overload << for pair so we can print pairs if needed
template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second;
}

// Overload >> for pair so we can read pairs directly
template<typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& x) {
    return in >> x.first >> x.second;
}

// Overload >> for vector to read all its elements
template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    for (auto& x: a) {
        in >> x;
    }
    return in;
}

// Overload << for vector to print its elements
template<typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
    for (auto x: a) {
        out << x << ' ';
    }
    return out;
}

int n, m;            // Number of vertices and edges
vector<int> deg;     // deg[i] will hold the degree of vertex i

// Read input: N, M, followed by M pairs of endpoints
void read() {
    cin >> n >> m;
    deg.assign(n, 0);    // Initialize all degrees to 0
    for (int i = 0; i < m; i++) {
        int u, v;
        cin >> u >> v;
        --u;               // convert to 0-based index
        --v;
        deg[u]++;          // increment degree of u
        deg[v]++;          // increment degree of v
    }
}

// Compute and print the answer
void solve() {
    long long ans = 0;
    // Sum over vertices: d_v * (d_v - 1)
    for (int i = 0; i < n; i++) {
        ans += 1LL * deg[i] * (deg[i] - 1);
    }
    // Add 2*M for the diagonal entries of A^T A
    ans += 2LL * m;
    cout << ans << "\n";
}

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

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

4. Python Solution with Detailed Comments  
```python
import sys
data = sys.stdin.read().split()
it = iter(data)

# Read N and M
n = int(next(it))
m = int(next(it))

# Initialize degree list of size n to zeros
deg = [0] * n

# Read each edge, convert to 0-based indices, and update degrees
for _ in range(m):
    u = int(next(it)) - 1
    v = int(next(it)) - 1
    deg[u] += 1
    deg[v] += 1

# Compute the contribution from ordered pairs of distinct edges sharing a vertex
pairs_sum = 0
for d in deg:
    # For vertex of degree d, there are d*(d-1) ordered pairs of incident edges
    pairs_sum += d * (d - 1)

# Add 2*M for the diagonal entries (each edge contributes 2 to its own dot product)
result = pairs_sum + 2 * m

# Print the final answer
print(result)
```

5. Compressed Editorial  
We want the sum of all entries of AᵀA, where A is the N×M incidence matrix. Each diagonal entry contributes 2, summing to 2M. Each off-diagonal entry (j≠k) is 1 iff edges j and k share a vertex; counting ordered pairs of edges around each vertex v of degree d_v gives d_v·(d_v−1). Total = 2M + ∑_v d_v(d_v−1). Compute degrees in O(M), sum over vertices in O(N).