## 1. Abridged problem statement

You are given `c` colors of 3D “corner” pieces. A corner of size `2^j` consists of 4 cubes of side `2^j`. You need to build one big corner of size `2^k`.

For each color `i` and size index `j`, you are given `a[i][j]`, the number of available corners of that color and size `2^j`, where `0 <= j < k`.

You must output a multiset of available pieces that can build the big corner, using the minimum possible number of different colors. If multiple answers exist, output any. If impossible, output `NO SOLUTION`.

Constraints:

```text
1 <= c, k <= 1000
0 <= a[i][j] <= 1000
```

---

## 2. Detailed editorial

### Key geometric observation

A corner of size `2^j` has volume proportional to:

```text
4 * (2^j)^3 = 4 * 8^j
```

The constant factor `4` is common for all corners, so we can measure a corner of size `2^j` as having “weight”:

```text
8^j
```

A corner of size `2^j` can be subdivided into exactly `8` corners of size `2^(j-1)`.

Therefore, to build one big corner of size `2^k`, we need total normalized volume:

```text
8^k
```

using pieces of sizes `0, 1, ..., k - 1`.

So if we choose some set of colors `S`, the necessary and sufficient condition for being able to build the big corner is:

```text
sum over colors i in S, sizes j of a[i][j] * 8^j >= 8^k
```

Why sufficient? Because any missing larger corner can always be replaced by `8` smaller corners recursively.

---

### Minimizing the number of colors

For each color, compute its total normalized volume:

```text
volume[i] = sum over j of a[i][j] * 8^j
```

Since feasibility depends only on total volume of chosen colors, to minimize the number of colors we should take the colors with largest total volume first.

So:

1. Compute total volume for each color.
2. Sort colors in decreasing order of total volume.
3. Add colors one by one until the chosen colors can build the big corner.

The first prefix that works uses the minimum possible number of colors.

Because `k` can be up to `1000`, values like `8^1000` are enormous. The C++ solution stores each volume as a base-8 digit vector and compares digit vectors instead of using big integers.

---

### Finding exact counts of pieces to use

After choosing a set of colors, we need to output how many pieces of each size and color are used.

Let:

```text
sum_caps[j] = total number of available chosen pieces of size 2^j
```

We greedily decide how many pieces of each size to use, starting from the largest size `k - 1`.

Initially, a size-`2^k` corner can be made from:

```text
8
```

corners of size `2^(k-1)`.

So set:

```text
need = 8
```

Then process sizes from `k - 1` down to `0`.

At size `j`:

- If we have at least `need` pieces of size `j`, use exactly `need` and finish.
- Otherwise, use all available pieces of size `j`.
- Every missing size-`j` piece must be replaced by `8` pieces of size `j - 1`.

So:

```text
missing = need - sum_caps[j]
need = missing * 8
```

If after processing size `0` we still cannot satisfy the demand, construction is impossible with the chosen colors.

---

### Producing the output

Once we know the total number `x[j]` of pieces of size `j` to use, we distribute these counts among the chosen colors.

For each size `j`, iterate over the chosen colors and take as many pieces as possible until `x[j]` becomes zero.

Each output line is:

```text
color size count
```

Colors are 1-indexed in the output.

---

### Complexity

Let `c` be the number of colors and `k` the number of size levels.

Computing volumes:

```text
O(c * k)
```

Sorting colors:

```text
O(c log c * k)
```

because comparing two base-8 digit vectors may take `O(k)`.

Trying prefixes and greedy feasibility:

```text
O(c * k)
```

Total complexity:

```text
O(c * k + c log c * k)
```

This is fine for `c, k <= 1000`.

---

## 3. Provided C++ solution with detailed comments

```cpp
#include <bits/stdc++.h> // Includes almost all standard C++ headers.

using namespace std; // Allows using standard library names without std:: prefix.

// Output operator for pairs.
// Prints pair as: first second
template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second;
}

// Input operator for pairs.
// Reads pair as: first second
template<typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& x) {
    return in >> x.first >> x.second;
}

// Input operator for vectors.
// Reads all elements of the vector in order.
template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    for(auto& x: a) {
        in >> x;
    }
    return in;
};

// Output operator for vectors.
// Prints elements separated by spaces.
template<typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
    for(auto x: a) {
        out << x << ' ';
    }
    return out;
};

// c = number of colors.
// k = target corner size index. We need to build a corner of size 2^k.
int c, k;

// a[i][j] = number of corners of color i and size 2^j.
vector<vector<int>> a;

// Reads the input.
void read() {
    cin >> c >> k; // Read number of colors and size index of big corner.

    // Allocate c rows and k columns.
    a.assign(c, vector<int>(k));

    // Read all available piece counts.
    for(int i = 0; i < c; i++) {
        for(int j = 0; j < k; j++) {
            cin >> a[i][j];
        }
    }
}

// Converts coefficients of powers of 8 into normalized base-8 digits.
//
// Input coeffs represent:
//
//   coeffs[0] * 8^0 + coeffs[1] * 8^1 + ...
//
// But coeffs[j] may be larger than 7.
// This function performs carries so every digit is in [0, 7].
vector<int> normalize_base8(const vector<int>& coeffs) {
    vector<int> digits; // Normalized base-8 digits, least significant first.

    // Reserve a little extra space in case carries create new digits.
    digits.reserve(coeffs.size() + 4);

    int64_t carry = 0; // Carry from lower base-8 positions.

    // Process every coefficient.
    for(int v: coeffs) {
        // Add carry from previous lower digit.
        int64_t cur = (int64_t)v + carry;

        // Current normalized base-8 digit.
        digits.push_back((int)(cur % 8));

        // Carry to the next digit.
        carry = cur / 8;
    }

    // Continue processing remaining carry.
    while(carry > 0) {
        digits.push_back((int)(carry % 8));
        carry /= 8;
    }

    // Remove leading zero digits.
    while(!digits.empty() && digits.back() == 0) {
        digits.pop_back();
    }

    // Return normalized base-8 representation.
    return digits;
}

// Compares two base-8 digit vectors.
//
// Digits are stored least significant first.
// Returns:
//   -1 if x < y
//    0 if x == y
//    1 if x > y
int cmp_digits(const vector<int>& x, const vector<int>& y) {
    // A number with more base-8 digits is larger.
    if(x.size() != y.size()) {
        return x.size() < y.size() ? -1 : 1;
    }

    // Same number of digits: compare from most significant digit down.
    for(int i = (int)x.size() - 1; i >= 0; i--) {
        if(x[i] != y[i]) {
            return x[i] < y[i] ? -1 : 1;
        }
    }

    // All digits equal.
    return 0;
}

// Main solving function.
void solve() {
    /*
        A corner of size 2^j has normalized volume 8^j.

        The big corner has size 2^k, so its normalized volume is 8^k.

        A corner of size 2^j can be split into 8 corners of size 2^(j-1).
        Therefore, construction is possible using a chosen set of colors iff
        their total normalized volume is at least 8^k.

        To minimize the number of colors, sort colors by total volume descending
        and take the shortest prefix whose total volume is enough.

        After choosing colors, greedily determine how many pieces of each size
        to use, starting from the largest size.
    */

    // vol_digits[i] stores total volume of color i in normalized base-8 form.
    vector<vector<int>> vol_digits(c);

    // Compute normalized base-8 volume representation for every color.
    for(int i = 0; i < c; i++) {
        vol_digits[i] = normalize_base8(a[i]);
    }

    // order will contain color indices sorted by decreasing volume.
    vector<int> order(c);

    // Fill order with 0, 1, ..., c - 1.
    iota(order.begin(), order.end(), 0);

    // Sort colors by descending total volume.
    sort(order.begin(), order.end(), [&](int u, int v) {
        return cmp_digits(vol_digits[u], vol_digits[v]) > 0;
    });

    // sum_caps[j] = total available pieces of size 2^j among chosen colors.
    vector<int64_t> sum_caps(k, 0);

    // x[j] = number of pieces of size 2^j that we will actually use.
    vector<int64_t> x(k, 0);

    // List of chosen color indices.
    vector<int> chosen;

    // Whether we have found a feasible prefix of colors.
    bool feasible = false;

    // Lambda that tests if currently chosen colors can build the big corner.
    auto try_greedy = [&]() -> bool {
        // Reset selected counts for each size.
        fill(x.begin(), x.end(), (int64_t)0);

        // To build one size-2^k corner, initially we need 8 corners of size 2^(k-1).
        int64_t N = 8;

        // Clamp value to avoid overflow.
        //
        // Each sum_caps[j] is at most c * 1000 <= 1,000,000.
        // Once N becomes very large, it can never be satisfied later.
        const int64_t CLAMP = (int64_t)4e9;

        // Process sizes from largest to smallest.
        for(int j = k - 1; j >= 0; j--) {
            // If we have enough corners of this size, use exactly N and finish.
            if(sum_caps[j] >= N) {
                x[j] = N;
                N = 0;
                break;
            }

            // Otherwise, use all available corners of this size.
            x[j] = sum_caps[j];

            // Remaining corners of this size that must be replaced by smaller ones.
            int64_t rem = N - sum_caps[j];

            // Avoid overflow when multiplying by 8.
            if(rem > CLAMP / 8) {
                N = CLAMP;
            } else {
                // Each missing size-j corner requires 8 size-(j-1) corners.
                N = rem * 8;
            }
        }

        // Feasible iff all required volume was satisfied.
        return N == 0;
    };

    // Add colors in descending volume order until construction becomes feasible.
    for(int idx = 0; idx < c; idx++) {
        // Current color index.
        int cc = order[idx];

        // Mark this color as chosen.
        chosen.push_back(cc);

        // Add this color's pieces to total capacities.
        for(int j = 0; j < k; j++) {
            sum_caps[j] += a[cc][j];
        }

        // Test whether the chosen prefix can build the big corner.
        if(try_greedy()) {
            feasible = true;
            break;
        }
    }

    // If even all colors are not enough, there is no solution.
    if(!feasible) {
        cout << "NO SOLUTION\n";
        return;
    }

    // result stores triples:
    //   color index, size index, count
    vector<tuple<int, int, int>> result;

    // Distribute required counts x[j] among chosen colors.
    for(int j = 0; j < k; j++) {
        // Number of size-2^j pieces still needed.
        int64_t need = x[j];

        // Try to take them from chosen colors.
        for(int cc: chosen) {
            // Stop once this size is fully supplied.
            if(need == 0) {
                break;
            }

            // Take as many as possible from this color and size.
            int64_t take = min((int64_t)a[cc][j], need);

            // If we take at least one piece, output this type.
            if(take > 0) {
                // Colors are output 1-indexed, size remains j.
                result.push_back({cc + 1, j, (int)take});

                // Decrease remaining demand.
                need -= take;
            }
        }
    }

    // Print number of used piece types.
    cout << result.size() << "\n";

    // Print each used type.
    for(auto& [col, sz, cnt]: result) {
        cout << col << ' ' << sz << ' ' << cnt << "\n";
    }
}

// Program entry point.
int main() {
    // Fast input/output.
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

    // The problem has one test case.
    int T = 1;

    // Multiple test cases are not used.
    // cin >> T;

    // Solve each test case.
    for(int test = 1; test <= T; test++) {
        read();
        solve();
    }

    return 0;
}
```

---

## 4. Python solution with detailed comments

```python
import sys


def solve():
    # Read all integers from standard input.
    data = list(map(int, sys.stdin.buffer.read().split()))

    # If input is empty, do nothing.
    if not data:
        return

    # Pointer into the flat input array.
    ptr = 0

    # Number of colors.
    c = data[ptr]
    ptr += 1

    # Big corner has size 2^k.
    k = data[ptr]
    ptr += 1

    # a[i][j] = number of available corners of color i and size 2^j.
    a = []

    # Read the c by k table.
    for _ in range(c):
        row = data[ptr:ptr + k]
        ptr += k
        a.append(row)

    # Compute normalized volume of each color.
    #
    # A corner of size 2^j has normalized volume 8^j.
    # Python integers are arbitrary precision, so we can store huge values directly.
    volumes = []

    for i in range(c):
        total = 0

        for j in range(k):
            cnt = a[i][j]

            # cnt * 8^j.
            #
            # Since 8^j == 2^(3j), shifting left by 3*j is equivalent.
            total += cnt << (3 * j)

        volumes.append(total)

    # Sort color indices by decreasing total volume.
    order = sorted(range(c), key=lambda i: volumes[i], reverse=True)

    # sum_caps[j] = total available pieces of size 2^j among chosen colors.
    sum_caps = [0] * k

    # x[j] = number of pieces of size 2^j that we will actually use.
    x = [0] * k

    # Chosen color indices.
    chosen = []

    def try_greedy():
        """
        Tests whether currently chosen colors can build the big corner.

        Also fills array x, where x[j] is the number of pieces of size 2^j
        that should be used.

        We process sizes from large to small.

        To build a size-2^k corner, we initially need 8 corners of size 2^(k-1).
        If some of those are missing, each missing one is replaced by 8 smaller
        corners.
        """

        # We want to modify outer variable x.
        nonlocal x

        # Reset chosen counts.
        x = [0] * k

        # Initial demand at size k - 1.
        need = 8

        # Process sizes from k - 1 down to 0.
        for j in range(k - 1, -1, -1):
            # If enough pieces of this size exist, use exactly need and finish.
            if sum_caps[j] >= need:
                x[j] = need
                return True

            # Otherwise, use all available pieces of this size.
            x[j] = sum_caps[j]

            # Missing pieces of size j.
            missing = need - sum_caps[j]

            # Each missing size-j piece requires 8 size-(j-1) pieces.
            need = missing * 8

        # If demand remains after size 0, construction is impossible.
        return False

    # Whether a feasible prefix of colors has been found.
    feasible = False

    # Add colors from largest volume to smallest.
    for color in order:
        # Add this color to the chosen set.
        chosen.append(color)

        # Add this color's pieces to total capacities.
        for j in range(k):
            sum_caps[j] += a[color][j]

        # Check if the current chosen colors are sufficient.
        if try_greedy():
            feasible = True
            break

    # If even all colors are insufficient, output NO SOLUTION.
    if not feasible:
        sys.stdout.write("NO SOLUTION\n")
        return

    # Store output triples:
    #   color, size, count
    result = []

    # For every size, distribute the required count among chosen colors.
    for j in range(k):
        # Number of size-2^j pieces still needed.
        need = x[j]

        # Take pieces from chosen colors.
        for color in chosen:
            # If enough pieces of this size have been taken, stop.
            if need == 0:
                break

            # Number of pieces we can take from this color and size.
            take = min(a[color][j], need)

            # If positive, record this type.
            if take > 0:
                # Output colors are 1-indexed.
                result.append((color + 1, j, take))

                # Decrease remaining demand.
                need -= take

    # Prepare output lines.
    out = []

    # First line: number of used types.
    out.append(str(len(result)))

    # Then each used type.
    for color, size, count in result:
        out.append(f"{color} {size} {count}")

    # Print answer.
    sys.stdout.write("\n".join(out) + "\n")


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

---

## 5. Compressed editorial

A corner of size `2^j` has volume proportional to `8^j`. A corner of size `2^j` can be replaced by `8` corners of size `2^(j-1)`. Therefore, for any chosen set of colors, construction is possible iff their total normalized volume is at least `8^k`.

For each color compute:

```text
volume[i] = sum_j a[i][j] * 8^j
```

To minimize the number of colors, sort colors by decreasing `volume[i]` and take the shortest prefix with enough volume.

Then determine exact piece counts greedily from large sizes to small sizes. Initially need `8` pieces of size `k - 1`. At size `j`, if available pieces are enough, use the needed amount and stop. Otherwise, use all available pieces and replace each missing piece by `8` smaller pieces.

Finally distribute the required counts among the chosen colors and output the used `(color, size, count)` triples. If no prefix works, output `NO SOLUTION`.