## 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. 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 c, k;
vector<vector<int>> a;

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

vector<int> normalize_base8(const vector<int>& coeffs) {
    vector<int> digits;
    digits.reserve(coeffs.size() + 4);
    int64_t carry = 0;
    for(int v: coeffs) {
        int64_t cur = (int64_t)v + carry;
        digits.push_back((int)(cur % 8));
        carry = cur / 8;
    }

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

    while(!digits.empty() && digits.back() == 0) {
        digits.pop_back();
    }

    return digits;
}

int cmp_digits(const vector<int>& x, const vector<int>& y) {
    if(x.size() != y.size()) {
        return x.size() < y.size() ? -1 : 1;
    }

    for(int i = (int)x.size() - 1; i >= 0; i--) {
        if(x[i] != y[i]) {
            return x[i] < y[i] ? -1 : 1;
        }
    }

    return 0;
}

void solve() {
    // A "corner" of size 2^j sits inside a bounding cube of side 2^{j+1} and
    // occupies 4 of the 8 unit sub-cubes meeting at one of its corners. The
    // opposite four sub-cubes form another corner, so two corners of size 2^j
    // tile a cube of side 2^{j+1}. Hence the big corner of size 2^k splits
    // into 4 cubes of side 2^k = 8 corners of size 2^{k-1}, and recursively
    // any corner of size 2^j can be replaced by 8 corners of size 2^{j-1}.
    //
    // Because of that recursion, the only constraint on which multiset of
    // small corners fills the big corner is the total volume: if x_j corners
    // of size 2^j are chosen with sum x_j * 8^j = 8^k, the tiling is always
    // realizable. The per-level cap x_j <= N_j (demand cascading from above)
    // is automatic, since sum_{i >= j} x_i * 8^i <= 8^k forces x_j to fit
    // the remaining demand. So feasibility for a chosen set S of colors is
    // just sum_{c in S, j} a[c][j] * 8^j >= 8^k.
    //
    // Minimizing the number of colors is then "smallest prefix of largest
    // volumes that reaches the target". Volumes are up to 8^1000, so each
    // color's volume is stored as a base-8 digit vector (after normalizing
    // a[c][.] by carry propagation), and colors are sorted by lexicographic
    // comparison of those digit vectors. We add colors one at a time in that
    // order and after each addition run the top-down greedy on the combined
    // per-size capacities T_j to test feasibility.
    //
    // The top-down greedy itself uses 64-bit arithmetic with a clamp on the
    // unmet demand N: as long as N stays below ~4e9 we propagate it
    // exactly; once it exceeds that bound it can no longer be absorbed by
    // any T_j (capped at c * 1000 = 1e6), so we clamp and treat the run as
    // infeasible. When feasibility holds, the loop terminates at the first
    // level where T_j >= N_j, giving the totals x_j; these are then split
    // across the chosen colors in any order, respecting a[c][j].

    vector<vector<int>> vol_digits(c);
    for(int i = 0; i < c; i++) {
        vol_digits[i] = normalize_base8(a[i]);
    }

    vector<int> order(c);
    iota(order.begin(), order.end(), 0);
    sort(order.begin(), order.end(), [&](int u, int v) {
        return cmp_digits(vol_digits[u], vol_digits[v]) > 0;
    });

    vector<int64_t> sum_caps(k, 0);
    vector<int64_t> x(k, 0);
    vector<int> chosen;
    bool feasible = false;

    auto try_greedy = [&]() -> bool {
        fill(x.begin(), x.end(), (int64_t)0);
        int64_t N = 8;
        const int64_t CLAMP = (int64_t)4e9;
        for(int j = k - 1; j >= 0; j--) {
            if(sum_caps[j] >= N) {
                x[j] = N;
                N = 0;
                break;
            }

            x[j] = sum_caps[j];
            int64_t rem = N - sum_caps[j];
            if(rem > CLAMP / 8) {
                N = CLAMP;
            } else {
                N = rem * 8;
            }
        }
        return N == 0;
    };

    for(int idx = 0; idx < c; idx++) {
        int cc = order[idx];
        chosen.push_back(cc);
        for(int j = 0; j < k; j++) {
            sum_caps[j] += a[cc][j];
        }

        if(try_greedy()) {
            feasible = true;
            break;
        }
    }

    if(!feasible) {
        cout << "NO SOLUTION\n";
        return;
    }

    vector<tuple<int, int, int>> result;
    for(int j = 0; j < k; j++) {
        int64_t need = x[j];
        for(int cc: chosen) {
            if(need == 0) {
                break;
            }

            int64_t take = min((int64_t)a[cc][j], need);
            if(take > 0) {
                result.push_back({cc + 1, j, (int)take});
                need -= take;
            }
        }
    }

    cout << result.size() << "\n";
    for(auto& [col, sz, cnt]: result) {
        cout << col << ' ' << sz << ' ' << cnt << "\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();
        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`.
