1. Abridged Problem Statement
Given an N×N grid with N≤3, find nonnegative integer values A[i][j] (0–9) so that for each cell (i,j), exactly B[i][j] of its up/down/left/right neighbors have strictly larger A-values. If a solution exists, output any; otherwise print “NO SOLUTION.”

2. Detailed Editorial

We have a tiny grid (at most 3×3), so we can brute-force the assignment of A[i][j] from 0 to 9 with backtracking. We also maintain a working copy of B, called b2, which tracks how many larger neighbors are still “needed” for each cell as we assign values.

Key ideas:
- When we assign a value v to cell (i,j), we affect the neighbor-relation count for its up and left neighbors (since those are already assigned if we fill row by row).
- For each adjacent cell (x,y) already assigned, if A[x][y]>v we decrement b2[x][y] by 1 (one fewer larger neighbor needed), and if v>A[x][y] we decrement b2[i][j] by 1.
- We never allow any b2 entry to go negative.
- After assigning all cells, we check that the bottom row’s b2 entries are zero; combined with the earlier checks (each filled cell requires its upper neighbor's residual to hit zero before recursing), this guarantees every cell’s b2 is zero, so every B is satisfied.
- If at any point an entry goes negative or we cannot satisfy a previously filled neighbor’s count, we backtrack.

Because N≤3 and each cell has only 10 possible values, this DFS tries at most 10^(N²)=10⁹ in the very worst theoretical bound, but in practice the pruning from b2 failing keeps it extremely fast, and on N≤3 it runs instantly.

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;
};

int n;
vector<vector<int>> a, b, b2;

void read() {
    cin >> n;
    b.resize(n, vector<int>(n));
    a.resize(n, vector<int>(n));
    cin >> b;
    b2 = b;
}

bool change(int i, int j, int delta) {
    bool fail = false;
    if(i > 0) {
        b2[i][j] += delta * (a[i - 1][j] > a[i][j]);
        b2[i - 1][j] += delta * (a[i - 1][j] < a[i][j]);
        if(b2[i][j] < 0 || b2[i - 1][j] < 0) {
            fail = true;
        }
    }

    if(j > 0) {
        b2[i][j] += delta * (a[i][j - 1] > a[i][j]);
        b2[i][j - 1] += delta * (a[i][j - 1] < a[i][j]);
        if(b2[i][j] < 0 || b2[i][j - 1] < 0) {
            fail = true;
        }
    }
    return fail;
}

void backtrack(int pos) {
    if(pos == n * n) {
        for(int j = 0; j < n; j++) {
            if(b2[n - 1][j] != 0) {
                return;
            }
        }

        for(int i = 0; i < n; i++) {
            for(int j = 0; j < n; j++) {
                cout << a[i][j] << " \n"[j == n - 1];
            }
        }
        exit(0);
    }

    int i = pos / n, j = pos % n;
    for(int val = 0; val <= 9; val++) {
        a[i][j] = val;
        if(!change(i, j, -1) && (i == 0 || b2[i - 1][j] == 0)) {
            backtrack(pos + 1);
        }
        change(i, j, 1);
    }
}

void solve() {
    /*
     * Reconstruct the army matrix A by backtracking over cells in row-major
     * order, trying every digit 0..9 per cell.
     *
     * b2 is a working copy of the target neighbour-counts B that is decremented
     * as constraints get satisfied: change(i, j, delta) adjusts the counts of
     * the current cell and its already-placed up/left neighbours according to
     * which side holds the larger army. After filling a cell we require its
     * upper neighbour's residual count to be exactly zero (it can no longer
     * gain neighbours), pruning impossible branches. A full assignment whose
     * last row also has all-zero residuals is printed and the program exits.
     */

    backtrack(0);
    cout << "NO SOLUTION" << '\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
import sys
sys.setrecursionlimit(10000)

def read_input():
    data = sys.stdin.read().strip().split()
    n = int(data[0])
    B = []
    idx = 1
    for _ in range(n):
        row = list(map(int, data[idx:idx+n]))
        idx += n
        B.append(row)
    return n, B

def solve():
    n, B = read_input()
    # A: grid we build; b2: working neighbor counts
    A = [[0]*n for _ in range(n)]
    b2 = [row[:] for row in B]

    # Update b2 when placing/removing A[i][j]
    # delta = -1 to place, +1 to remove
    def update(i, j, delta):
        # Returns False if any count goes negative
        ok = True
        for di, dj in [(-1,0),(0,-1)]:
            x, y = i+di, j+dj
            if 0 <= x < n and 0 <= y < n:
                # If neighbor > this, adjust this cell’s count
                if A[x][y] > A[i][j]:
                    b2[i][j] += delta
                # If this > neighbor, adjust neighbor’s count
                if A[i][j] > A[x][y]:
                    b2[x][y] += delta
                if b2[i][j] < 0 or b2[x][y] < 0:
                    ok = False
        return ok

    # DFS over flat positions
    def dfs(pos):
        if pos == n*n:
            # Check last row all zero
            if all(b2[n-1][j]==0 for j in range(n)):
                for row in A:
                    print(" ".join(map(str,row)))
                sys.exit(0)
            return

        i, j = divmod(pos, n)
        for val in range(10):
            A[i][j] = val
            if update(i, j, -1):
                # Ensure that the above cell (if exists) is settled
                if i==0 or b2[i-1][j]==0:
                    dfs(pos+1)
            update(i, j, +1)

    dfs(0)
    print("NO SOLUTION")

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

5. Compressed Editorial
Use DFS to assign each A[i][j] in 0–9, tracking how many larger neighbors remain needed in a working matrix b2. On each placement/removal, update b2 for the up and left neighbors only and backtrack on any negative counts. Upon filling all cells, verify the last row’s counts are zero; if so, print the grid.
