1. Abridged Problem Statement  
Given an N×N grid of cells numbered 1…N² in row-major order, you start with a pointer on cell 1. You must plan a sequence of turns. On turn i you choose a distinct Ki (N ≤ Ki < 300) and tell the audience to move the pointer Ki steps (each step to an adjacent cell). Then you remove at least one cell—none of which may be the current pointer cell. After e turns, exactly one cell must remain. Output each turn as:  
Ki Xi,1 Xi,2 … Xi,mi  

2. Detailed Editorial  

Goal  
We must guarantee that whenever we remove a set of cells after Ki moves, the audience’s unknown position cannot lie in the removed set. We do this by exploiting two well-known facts about reachability on a grid:

1. From a starting cell, the cells reachable in exactly K steps are those whose Manhattan distance d from the start satisfies d ≤ K and (K−d) even.  
2. Parity (even/odd) of distance matters: you cannot reach a cell of opposite color on a checkerboard in an even number of steps, and vice versa.

Construction Overview  
We remove cells in “layers” defined by anti-diagonals (cells with constant i+j).  Index rows and columns 0…N−1, start at (0,0).  Let sum = i+j.

A) First Turn (remove everything with sum > N):  
- Choose K1 = N.  
- Reachable cells in exactly N steps from (0,0) are those with sum ≤ N and sum≡N (mod 2).  
- Any cell with sum > N is unreachable ⇒ safe to remove all of them in one shot.

B) Remaining Possible Positions S1  
After removing sum>N, the pointer must lie in S1 = {cells with sum≤N and parity(sum)=parity(N)}.

C) Removing One Diagonal Per Turn  
We will now remove all cells with sum=N, then sum=N−1, …, down to sum=1, one diagonal each turn.  To eliminate the diagonal sum=d, we pick Ki that makes every cell on that diagonal unreachable from S1 in exactly Ki moves.

- Notice all cells in S1 have parity(sum)=parity(N).  
- If we pick Ki to be an odd number (and larger than N), then after Ki moves the pointer must land on a cell of opposite parity to parity(N).  
- But the diagonal with sum=d has parity(sum)=d≡parity(N) when d and N have same parity.  By choosing diagonals in decreasing order N, N−1, …,1, we ensure each targeted diagonal has parity = parity(N) on every odd-numbered removal.  Hence those cells become unreachable ⇒ safe to remove that entire diagonal.

D) Ki Selection and Turn Count  
- K1 = N (removes sum>N).  
- For the next N turns (to kill diagonals from sum=N down to sum=1), choose Ki = the smallest odd number ≥N+1, then increase by 2 each time to keep them distinct.  
- After these N+1 total turns, only the cell (0,0) with sum=0 remains.

Complexity  
We loop over all N² cells once for the first removal, and then over N diagonals of total size N each. Overall O(N²), which is fine for N≤100.

3. Provided C++ Solution with Detailed Comments  

```cpp
#include <bits/stdc++.h>
using namespace std;

// Overload << for pair printing (not used in this problem but included).
template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second;
}

// Overload >> for pair reading.
template<typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& x) {
    return in >> x.first >> x.second;
}

// Overload >> for vector reading.
template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    for(auto &x: a) in >> x;
    return in;
}

// Overload << for vector printing.
template<typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
    for(auto x: a) out << x << ' ';
    return out;
}

int n;

// Read input N.
void read() {
    cin >> n;
}

void solve() {
    // ans will store each line: first element is Ki, rest are cells to remove.
    vector<vector<int>> ans;

    // ---- Turn 1: Ki = n, remove all cells with i+j > n ----
    // Start a new line with Ki = n.
    ans.push_back({n});
    // Loop over all grid positions (i,j), 0-based.
    // If i + j > n, compute its 1-based ID and remove it.
    for(int i = 0; i < n; i++) {
        for(int j = 0; j < n; j++) {
            if(i + j > n) {
                int id = i * n + j + 1;  // Convert to 1..n^2
                ans.back().push_back(id);
            }
        }
    }

    // ---- Subsequent N turns: remove diagonals sum=N, N-1, ..., 1 ----
    // We'll use odd Ki values starting from the smallest odd ≥ n+1.
    int q = n + 1;
    if(q % 2 == 0) q++;  // make q odd if needed

    // steps = 0 removes diagonal sum = n, steps=1 sum = n-1, ..., steps=n-1 sum=1
    for(int steps = 0; steps < n; steps++) {
        // Begin a new turn with Ki = q
        ans.push_back({q});

        // Compute starting point (diag_x, diag_y) to traverse the diagonal with sum = n - steps.
        int diag_x = 0, diag_y = n - steps;
        // Special case for the first diagonal: we skip the cell (0,n) which is out of grid.
        if(steps == 0) {
            diag_x = 1;
            diag_y = n - 1;
        }

        // Walk along the diagonal by increasing x and decreasing y.
        while(diag_y >= 0 && diag_x < n) {
            int id = diag_x * n + diag_y + 1;  // 1-based ID
            ans.back().push_back(id);
            diag_x++;
            diag_y--;
        }
        // Move to the next odd value for the next turn.
        q += 2;
    }

    // Output all turns.
    for(const auto &line : ans) {
        for(int x : line) {
            cout << x << ' ';
        }
        cout << '\n';
    }
}

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

    read();
    solve();
    return 0;
}
```

4. Python Solution with Detailed Comments  

```python
def magic_trick(N):
    # We'll build a list of turns. Each turn is [K, x1, x2, ...].
    turns = []

    # --- Turn 1 ---
    # K1 = N; remove all cells with i+j > N.
    line1 = [N]
    for i in range(N):
        for j in range(N):
            if i + j > N:
                # Compute 1-based index
                cell_id = i * N + j + 1
                line1.append(cell_id)
    turns.append(line1)

    # --- Subsequent N turns: remove diagonals ---
    # Start with the smallest odd K >= N+1
    K = N + 1
    if K % 2 == 0:
        K += 1

    # For each diagonal sum = N, N-1, ..., 1
    for step in range(N):
        current = [K]
        diag_sum = N - step

        # We want all (i, j) with i+j = diag_sum
        # But skip the invalid cell (0, N) on the first pass.
        if step == 0:
            i, j = 1, N-1
        else:
            i, j = 0, diag_sum

        # Walk the diagonal
        while 0 <= j < N and 0 <= i < N:
            cell_id = i * N + j + 1
            current.append(cell_id)
            i += 1
            j -= 1

        turns.append(current)
        K += 2  # next odd

    return turns

def main():
    import sys
    data = sys.stdin.read().strip().split()
    N = int(data[0])
    for line in magic_trick(N):
        print(' '.join(map(str, line)))

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

Explanation of Key Steps  
- First removal (K=N) clears all cells too far (i+j>N).  
- Then each further turn clears exactly one anti-diagonal (sum constant) by choosing an odd K large enough so that those diagonal cells are unreachable (parity mismatch) from any of the remaining positions.  
- After N+1 turns, only the (0,0) cell survives.

5. Compressed Editorial  
1. Reachability in exactly K moves ⇒ Manh. dist ≤K with parity match.  
2. Turn 1: K=N removes all cells with i+j>N (too far). Remaining positions all have parity i+j≡N.  
3. Next N turns: pick odd K>N. Each odd K flips parity, so cells on any diagonal with parity=N become unreachable ⇒ safe to remove whole diagonal.  
4. Remove diagonals sum=N…1 in order. Only sum=0 cell remains.