1. Abridged Problem Statement  
Given a 3×3 grid of integers, compute the median of each row (the middle value when that row’s three numbers are sorted), then output the median of those three medians.

2. Detailed Editorial  
Task restated  
- You have three triples of integers.  
- For each triple, find its median (the second-smallest of the three).  
- Finally, take the three medians you obtained and find their median.  

Key observations  
- Each row contains only three numbers, so finding a median is trivial.  
- We can sort each triple of size 3 in O(1) time (effectively a constant operation).  
- We then have three medians; again we can sort or manually pick the middle one.  
- Total work is constant time, well within any limits.

Possible approaches  
A. Full sort approach  
   1. Read each row into an array of size 3.  
   2. Sort it (e.g. with std::sort or any 3-element sort network).  
   3. Record the middle element as that row’s median.  
   4. Collect the three medians into another array of size 3, sort it, and take its middle element.  
B. Branch-based selection  
   - Since size = 3 always, you can use comparisons to pick the middle—there are only a few orderings.

Implementation details  
- Use fast I/O (although unnecessary for 9 integers, but idiomatic).  
- Pay attention to negative numbers (sorting handles that automatically).  
- Print the final result and exit.

Complexity  
- Time: O(1) (constant time operations on fixed size arrays).  
- Memory: O(1) extra.

3. C++ Solution
```cpp
#include <bits/stdc++.h>

using namespace std;

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;
};

vector<vector<int>> a;

void read() {
    a.assign(3, vector<int>(3));
    for(int i = 0; i < 3; i++) {
        cin >> a[i];
    }
}

void solve() {
    // Sort each of the three triples, take the middle element as that triple's
    // median, then sort those three medians and print the middle one.

    vector<int> medians;
    for(int i = 0; i < 3; i++) {
        sort(a[i].begin(), a[i].end());
        medians.push_back(a[i][1]);
    }

    sort(medians.begin(), medians.end());
    cout << medians[1] << '\n';
}

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

    int T = 1;
    // cin >> T;
    for(int test = 1; test <= T; test++) {
        read();
        // cout << "Case #" << test << ": ";
        solve();
    }

    return 0;
}
```

4. Python Solution with Detailed Comments  
```python
# Read three lines of input, each with three integers
grid = [list(map(int, input().split())) for _ in range(3)]

# For each row in the grid:
#  - Sort the row, then take row[1], the middle element
medians = []
for row in grid:
    row.sort()        # Sorts in non-decreasing order
    medians.append(row[1])

# Now we have three medians; sort them and pick the middle one
medians.sort()
print(medians[1])
```

5. Compressed Editorial  
Sort each of the three input rows of size 3 to get its middle value. Collect these three medians, sort them, and print their middle element. This runs in constant time.