1. Abridged Problem Statement  
Given integers N, M (3 ≤ N, M ≤ 200), paint an N×M grid with red (‘0’) and blue (‘#’) so that every 3×2 or 2×3 sub-rectangle contains exactly two blue cells. Among all valid paintings, use as few blue cells as possible. If there is no solution, print “No solution”; otherwise output any valid grid.

2. Detailed Editorial  

Goal: For every 3×2 or 2×3 block of cells, there must be exactly two ‘#’, and globally we want to minimize the total number of ‘#’.

Key Observation:  
If we paint exactly those cells (i,j) for which (i + j + offset) mod 3 = 0 as blue and all others red, then in any 3×2 or 2×3 block exactly two of the six cells satisfy (i + j) mod 3 = constant. Hence this pattern meets the local constraint. Why?  
- Consider any 3 consecutive rows and 2 consecutive columns: the six sums i+j form each residue class mod 3 exactly twice.  
- A similar argument holds for 2×3 blocks.

Thus the valid colorings are exactly the three “diagonal modulo-3” patterns, parameterized by offset ∈ {0,1,2}. Each has roughly ⌈N·M/3⌉ blue cells, but due to boundary effects one offset may use fewer blues. We simply try all three offsets, count the blues, pick the one with minimal count, and output that pattern.

Algorithm:  
1. Read N, M.  
2. For offset in {0,1,2}:  
   a. Count how many positions (i,j) with (i+j+offset)%3==0.  
3. Choose the offset that gives the smallest count.  
4. Construct the grid: mark ‘#’ at (i,j) satisfying (i+j+best_offset)%3==0, else ‘0’.  
5. Print the grid.

Time complexity: O(3·N·M) = O(N·M) which is fine for N,M ≤ 200.

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, m;

void read() { cin >> n >> m; }

void solve() {
    // Try all 3 possible diagonal patterns (offset 0, 1, 2)
    // Each pattern creates diagonal lines with period 3
    int min_count = INT_MAX;
    int best_offset = 0;

    // Test each of the 3 possible diagonal offsets
    for(int offset = 0; offset < 3; offset++) {
        int count = 0;

        // Count how many '#' symbols this offset would create
        for(int i = 0; i < n; i++) {
            // For row i, place '#' at columns: (i + offset) % 3, (i + offset) %
            // 3 + 3, (i + offset) % 3 + 6, ... This creates a diagonal pattern
            // with period 3
            for(int j = (i + offset) % 3; j < m; j += 3) {
                count++;
            }
        }

        // Keep track of the offset that minimizes the number of '#' symbols
        if(count < min_count) {
            min_count = count;
            best_offset = offset;
        }
    }

    // Create the grid using the best offset pattern
    // Use vector of strings for cleaner memory management
    vector<string> grid(n, string(m, '0'));

    // Fill the grid with '#' symbols using the optimal diagonal pattern
    for(int i = 0; i < n; i++) {
        // Place '#' at positions following the diagonal pattern
        for(int j = (i + best_offset) % 3; j < m; j += 3) {
            grid[i][j] = '#';
        }
    }

    // Output the resulting grid
    for(int i = 0; i < n; i++) {
        cout << grid[i] << endl;
    }
}

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
def main():
    import sys
    input = sys.stdin.readline

    # Read N, M
    n, m = map(int, input().split())

    best_offset = 0
    min_count = n * m + 1  # initialize to larger than maximum possible

    # Find which offset yields fewest blue cells
    for offset in range(3):
        cnt = 0
        for i in range(n):
            # Instead of checking every j, we can note that for fixed i,
            # j runs from 0 to m-1. We still check each one here for clarity.
            for j in range(m):
                if (i + j + offset) % 3 == 0:
                    cnt += 1
        if cnt < min_count:
            min_count = cnt
            best_offset = offset

    # Build the grid using the optimal offset
    grid = []
    for i in range(n):
        row = []
        for j in range(m):
            # Mark '#' if it satisfies the chosen diagonal pattern
            if (i + j + best_offset) % 3 == 0:
                row.append('#')
            else:
                row.append('0')
        grid.append(''.join(row))

    # Print the resulting flag
    print('\n'.join(grid))


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

5. Compressed Editorial  
Assign each cell (i,j) a value (i+j+offset)%3 and color it blue exactly when this value is 0. This guarantees every 3×2 or 2×3 block has exactly two blues (each residue appears twice), and by testing offsets 0,1,2 we pick the one minimizing blue count.