## 1) Concise abridged problem statement

You must simulate a simple text editor starting from an empty file (one empty line) with a cursor at line 0, column 0.  
Input is a single line of commands (max 20480 chars). Commands are:

- `a`–`z` or space: insert that character at the cursor position (shifting text right), then move cursor right.
- `L`: move cursor left within the line (not below 0).
- `R`: move cursor right (may go past end-of-line).
- `U` / `D`: move cursor up/down one line (staying within file).
- `N`: split line at cursor into two lines; tail goes to new line; cursor goes to start of new line.
- `E`: move cursor to end of current line.
- `H`: move cursor to start of current line.
- `X`: delete character at cursor; if cursor is past end-of-line, pad with spaces up to cursor and then merge next line into current (if exists).
- `B`: backspace (delete before cursor). If at col>0 delete previous; if at start of line, merge with previous line.
- `Q`: stop processing (ignore anything after).

Output the final text exactly: lines separated by newline, with **no extra newline after the last line**.

Constraints guarantee each line length ≤ 1000 at all times.

---

## 2) Detailed editorial (solution explanation)

### Data model
We store the file as:
- `vector<string> lines` — all lines of the editor.
- cursor position: `(row, col)` where:
  - `row` is index into `lines`
  - `col` is a cursor column that is allowed to be **greater** than `lines[row].size()` (cursor can be “after end-of-line”).

Initially:
- `lines = [""]`
- `row = 0`, `col = 0`

### Key helper: padding a line
Some operations require the line to exist up to the cursor column (notably inserting at `col` when `col > len`, and `X` when `col > len` and we need to merge lines).
So we use:

`pad_line(r, c)`:
- if `c > lines[r].size()`, resize the string to length `c` filling with spaces.

This matches the statement’s rule “additional spaces must be added” when editing beyond end-of-line.

### Processing commands
We scan the command string left to right, stopping at `Q`.

#### Inserting a lowercase letter or space
- Ensure the line has length at least `col` via `pad_line(row, col)`
- Insert character at position `col` (shifts right)
- `col++`

#### Cursor movement
- `L`: `col = max(0, col-1)`
- `R`: `col++` (allowed beyond end)
- `U`: if `row>0` then `row--`
- `D`: if `row+1 < lines.size()` then `row++`
No automatic column clamping occurs on vertical movement; that’s consistent with implementation and typical editor behavior. (Only `N` clamps before split.)

#### `N` (new line / split)
Before splitting, if cursor is beyond end-of-line, clamp it to end-of-line:
- `col = min(col, len)`

Then:
- `tail = line.substr(col)`
- current line becomes `line.substr(0, col)`
- insert `tail` as a new line after current
- move cursor to start of new line: `row++`, `col = 0`

#### `E` and `H`
- `E`: `col = lines[row].size()`
- `H`: `col = 0`

#### `X` (delete at cursor)
Case 1: `col < len`  
Delete that character: `erase(begin()+col)`

Case 2: `col >= len`
- If there is a next line (`row+1 < lines.size()`):
  - First pad current line to length `col` (adds spaces if needed)
  - Append next line contents to current line
  - Remove the next line from `lines`
- Else (last line): do nothing

This implements: deleting “current symbol” beyond end-of-line effectively joins the next line at the cursor position, but if cursor is beyond end you must insert spaces first.

#### `B` (backspace)
Case 1: `col > 0`
- If `col <= len`, delete character at `col-1` (normal backspace within text)
- If `col > len`, there is no character to delete (cursor is in virtual space), but the statement says cursor shifts left one position; the code does that by skipping erase and still doing `col--`.
- Finally `col--` always (when `col>0`)

Case 2: `col == 0`
- If `row > 0`, merge current line into previous:
  - remember `prev_len = lines[row-1].size()`
  - `lines[row-1] += lines[row]`
  - erase current line
  - `row--`, `col = prev_len` (cursor ends where the join occurred)
- If `row == 0`, do nothing

### Correctness argument (high level)
- Each command is implemented exactly per its definition, distinguishing cursor-inside-text vs cursor-beyond-end situations.
- `pad_line` ensures “additional spaces” are created precisely when required by insert and the special `X` join operation.
- Line splits and merges preserve all characters and relative order (split uses substrings; merge uses concatenation).
- Cursor updates match the spec: insert advances, `N` moves to new line start, joins place cursor correctly, etc.

### Complexity
Let `C` be number of commands (≤ 20480) and line lengths ≤ 1000.
Operations like `insert`/`erase` on `std::string` are O(length of line) due to shifting, but bounded by 1000, so worst-case time is about `O(C * 1000)` which fits easily.
Memory: total text size is bounded (lines lengths ≤ 1000, number of lines limited by operations), fits in 4MB.

---

## 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;
};

string commands;

void read() { getline(cin, commands); }

void solve() {
    // The hard part about this question is to understand what exactly the input
    // and output format is, otherwise it's just implementation.

    vector<string> lines = {""};
    int row = 0, col = 0;

    auto pad_line = [&](int r, int c) {
        if(c > (int)lines[r].size()) {
            lines[r].resize(c, ' ');
        }
    };

    for(int i = 0; i < (int)commands.size(); i++) {
        char c = commands[i];
        if(c == 'Q') {
            break;
        } else if(c >= 'a' && c <= 'z' || c == ' ') {
            pad_line(row, col);
            lines[row].insert(lines[row].begin() + col, c);
            col++;
        } else if(c == 'L') {
            if(col > 0) {
                col--;
            }
        } else if(c == 'R') {
            col++;
        } else if(c == 'U') {
            if(row > 0) {
                row--;
            }
        } else if(c == 'D') {
            if(row + 1 < (int)lines.size()) {
                row++;
            }
        } else if(c == 'N') {
            if(col > (int)lines[row].size()) {
                col = lines[row].size();
            }
            string tail = lines[row].substr(col);
            lines[row] = lines[row].substr(0, col);
            lines.insert(lines.begin() + row + 1, tail);
            row++;
            col = 0;
        } else if(c == 'E') {
            col = lines[row].size();
        } else if(c == 'H') {
            col = 0;
        } else if(c == 'X') {
            if(col < (int)lines[row].size()) {
                lines[row].erase(lines[row].begin() + col);
            } else if(row + 1 < (int)lines.size()) {
                pad_line(row, col);
                lines[row] += lines[row + 1];
                lines.erase(lines.begin() + row + 1);
            }
        } else if(c == 'B') {
            if(col > 0) {
                if(col <= (int)lines[row].size()) {
                    lines[row].erase(lines[row].begin() + col - 1);
                }
                col--;
            } else if(row > 0) {
                int prev_len = lines[row - 1].size();
                lines[row - 1] += lines[row];
                lines.erase(lines.begin() + row);
                row--;
                col = prev_len;
            }
        }
    }

    for(int i = 0; i < (int)lines.size(); i++) {
        if(i > 0) {
            cout << "\n";
        }
        cout << lines[i];
    }
}

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 (same behavior) with detailed comments

```python
import sys

def solve() -> None:
    # Read the entire command line including spaces, without the trailing newline.
    commands = sys.stdin.readline()
    if commands.endswith("\n"):
        commands = commands[:-1]

    # Editor state: list of lines and cursor (row, col).
    lines = [""]          # start with one empty line
    row, col = 0, 0

    def pad_line(r: int, c: int) -> None:
        """
        Ensure lines[r] has length at least c by padding with spaces.
        (Cursor column c is allowed to be beyond end-of-line.)
        """
        s = lines[r]
        if c > len(s):
            lines[r] = s + (" " * (c - len(s)))

    for ch in commands:
        if ch == "Q":
            break

        # Insert lowercase letters or spaces
        if ("a" <= ch <= "z") or ch == " ":
            pad_line(row, col)
            # Insert at position col: s[:col] + ch + s[col:]
            s = lines[row]
            lines[row] = s[:col] + ch + s[col:]
            col += 1

        elif ch == "L":
            if col > 0:
                col -= 1

        elif ch == "R":
            col += 1

        elif ch == "U":
            if row > 0:
                row -= 1

        elif ch == "D":
            if row + 1 < len(lines):
                row += 1

        elif ch == "N":
            # Clamp col to end-of-line before splitting (matches the C++ code).
            if col > len(lines[row]):
                col = len(lines[row])

            s = lines[row]
            head = s[:col]
            tail = s[col:]
            lines[row] = head
            lines.insert(row + 1, tail)
            row += 1
            col = 0

        elif ch == "E":
            col = len(lines[row])

        elif ch == "H":
            col = 0

        elif ch == "X":
            s = lines[row]
            if col < len(s):
                # Delete character at cursor.
                lines[row] = s[:col] + s[col + 1:]
            elif row + 1 < len(lines):
                # Beyond end-of-line: pad with spaces up to col, then merge next line.
                pad_line(row, col)
                lines[row] = lines[row] + lines[row + 1]
                del lines[row + 1]
            # Else: beyond end of last line => do nothing.

        elif ch == "B":
            if col > 0:
                s = lines[row]
                # If cursor is within the real text, delete the character before cursor.
                if col <= len(s):
                    lines[row] = s[:col - 1] + s[col:]
                # Cursor always moves left by one if col > 0.
                col -= 1
            elif row > 0:
                # Merge with previous line.
                prev_len = len(lines[row - 1])
                lines[row - 1] += lines[row]
                del lines[row]
                row -= 1
                col = prev_len
            # Else at (0,0): do nothing.

    # Output: newline between lines, but not after the last one.
    sys.stdout.write("\n".join(lines))

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

---

## 5) Compressed editorial

Simulate editor with `lines: list[str]` and cursor `(row,col)`; `col` may exceed current line length. Use `pad_line(row,col)` to extend a line with spaces to length `col` when needed.

Process commands until `Q`:
- insert letter/space: `pad_line`, insert at `col`, `col++`
- `L/R/U/D`: move cursor with bounds on `L/U/D`
- `H/E`: `col=0` / `col=len(line)`
- `N`: clamp `col` to end, split line into prefix+suffix, insert new line with suffix, go to `(row+1,0)`
- `X`: if `col<len` erase char; else if next line exists: `pad_line` then concatenate next line and remove it
- `B`: if `col>0` erase at `col-1` only if within text, then `col--`; else if `row>0` merge with previous and set `col` to previous length

Finally print all lines separated by `\n` without trailing newline. Complexity: O(commands * max_line_len) ≤ ~2e7 char ops, OK under constraints.