## 1) Abridged problem statement

You are given up to **N** (≤ 100) quoted strings (length 1…100) consisting only of **Latin letters**, **spaces**, and **'-'**. A *word* is a maximal sequence of letters.

You may **add/remove spaces** to produce a formatted version that obeys the ancient rules:

- Between two words with **no '-'** between them → exactly **one space**.
- A single `-` is a **hyphen** and must have a **word on both sides**: `word-word` (no spaces around it).
- A **short dash** is **two hyphens** with **spaces around**: `" -- "`.
- A **long dash** is **three hyphens** with **spaces around**: `" --- "`.

The output must contain **no other spaces** than those required by these rules.
If formatting is impossible, output `error` (without quotes). Otherwise print any valid formatted text **in quotes**.

---

## 2) Detailed editorial (solution explanation)

### Key observation
The only meaningful information in the input is:

- the sequence of **words** (letter-runs),
- and, between consecutive words (and also before the first / after the last), the **count of consecutive hyphens** `k` that appear (ignoring spaces).

Spaces in the input are irrelevant because we are allowed to add/remove them; hyphens are not removable, but may be regrouped into:
- hyphen between words (`k = 1` only allowed *between* two words),
- dash groups at boundaries or between words (`k ≥ 2`),
- or just normal word separation (`k = 0`).

So we parse the line into:
- `words = [w0, w1, ... w(n-1)]`
- `dashes = [d0, d1, ... dn]`
where:
- `d0` = number of `-` before `w0`
- `dj` (1 ≤ j ≤ n-1) = number of `-` between `w(j-1)` and `wj`
- `dn` = number of `-` after the last word

(Spaces are skipped while parsing.)

### Validity rules
1. Any character other than letter/space/`-` => impossible → `error`.
2. A single `-` (hyphen) must have a word on both sides.
   Therefore, `d0 != 1` and `dn != 1`. If either is 1 → `error`.
3. If there are **no words at all**:
   - If total hyphens is 1 → cannot be a hyphen (needs words) → `error`
   - If 0 → output empty string `""`
   - If ≥ 2 → it must be formatted purely as dashes with spaces according to dash rules.

### Constructing a formatted output
We now need to *print* the hyphen/dash groups using only valid patterns:

- `k = 0` between words → print `" "` (single space)
- `k = 1` between words → print `"-"`
- `k ≥ 2` → represent as a sequence of **short dashes** (`" -- "`) and at most one **long dash** (`" --- "`) at the front if needed, such that total hyphens sum to `k`.

This is always possible for `k ≥ 2`:
- if `k` is even: use only short dashes (2 hyphens each),
- if `k` is odd: use one long dash (3 hyphens) + remaining even part as short dashes.

Example: `k=7` → `" --- " + "-- " + "-- "` with correct spaces; the provided code builds a canonical form.

Finally:
- If `d0 > 0`, print its dash representation first,
- Then `w0`,
- For each next word `wj`: print separator based on `dj` then `wj`,
- If `dn > 0`, print its dash representation at the end.

All outputs are wrapped in quotes.

### Complexity
Each line is length ≤ 100, parsing and reconstruction are linear: **O(L)** time and O(L) memory, easily within limits.

---

## 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 text;

bool is_letter(char c) {
    return (c >= 'a' && c <= 'z') || (c >= 'A' && c <= 'Z');
}

void read() {
    string line;
    getline(cin, line);
    while(!line.empty() && (line.back() == '\r' || line.back() == '\n')) {
        line.pop_back();
    }
    if(line.size() >= 2 && line.front() == '"' && line.back() == '"') {
        text = line.substr(1, line.size() - 2);
    } else {
        text = "\x01";
    }
}

string dash_string(int k) {
    if(k == 0) {
        return "";
    }
    if(k == 1) {
        return "-";
    }

    string result;
    if(k % 2 == 1) {
        result = " --- ";
        k -= 3;
    } else {
        result = " -- ";
        k -= 2;
    }
    while(k > 0) {
        result += "-- ";
        k -= 2;
    }
    return result;
}

void solve() {
    // This is more of an exercise in parsing. There isn't really a complicated
    // algorithms idea here. We extract words and count dashes between them,
    // then validate that boundary dashes can be expressed as short/long dashes
    // (not single hyphens which require words on both sides).

    vector<string> words;
    vector<int> dashes;

    int dash_count = 0;
    int i = 0;
    while(i < (int)text.size()) {
        char c = text[i];
        if(c == ' ') {
            i++;
        } else if(c == '-') {
            dash_count++;
            i++;
        } else if(is_letter(c)) {
            dashes.push_back(dash_count);
            dash_count = 0;
            string word;
            while(i < (int)text.size() && is_letter(text[i])) {
                word += text[i];
                i++;
            }
            words.push_back(word);
        } else {
            cout << "error\n";
            return;
        }
    }
    dashes.push_back(dash_count);

    if(words.empty()) {
        if(dash_count == 1) {
            cout << "error\n";
        } else if(dash_count > 1) {
            cout << "\"" << dash_string(dash_count) << "\"\n";
        } else {
            cout << "\"\"\n";
        }
        return;
    }

    int n = words.size();
    if(dashes[0] == 1 || dashes[n] == 1) {
        cout << "error\n";
        return;
    }

    ostringstream oss;
    if(dashes[0] > 0) {
        oss << dash_string(dashes[0]);
    }
    oss << words[0];

    for(int j = 1; j < n; j++) {
        int k = dashes[j];
        if(k == 0) {
            oss << " ";
        } else if(k == 1) {
            oss << "-";
        } else {
            oss << dash_string(k);
        }
        oss << words[j];
    }

    if(dashes[n] > 0) {
        oss << dash_string(dashes[n]);
    }

    cout << "\"" << oss.str() << "\"\n";
}

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

    int T;
    cin >> T;
    cin.ignore(numeric_limits<streamsize>::max(), '\n');
    for(int test = 1; test <= T; test++) {
        read();
        solve();
    }

    return 0;
}
```

---

## 4) Python solution (same logic) with detailed comments

```python
import sys

def is_letter(c: str) -> bool:
    """Return True if c is an ASCII Latin letter."""
    return ('a' <= c <= 'z') or ('A' <= c <= 'Z')

def dash_string(k: int) -> str:
    """
    Convert k consecutive '-' into a valid dash representation.

    k=0 -> ""
    k=1 -> "-" (hyphen; only valid between two words)
    k>=2 -> combination of short dash " -- " (2 hyphens) and optionally one long dash " --- " (3 hyphens)
    """
    if k == 0:
        return ""
    if k == 1:
        return "-"

    # Build a canonical representation.
    # If k is odd, use one long dash first; else use one short dash first.
    if k % 2 == 1:
        res = " --- "
        k -= 3
    else:
        res = " -- "
        k -= 2

    # Remaining k is even; each short dash contributes 2 hyphens.
    while k > 0:
        res += "-- "
        k -= 2

    return res

def solve_one(line: str) -> str:
    """
    Solve for one quoted input line.
    Return either 'error' or a quoted formatted string.
    """
    line = line.rstrip("\r\n")

    # Validate and strip surrounding quotes
    if len(line) >= 2 and line[0] == '"' and line[-1] == '"':
        text = line[1:-1]
    else:
        # Invalid input format per statement; treat as impossible
        return "error"

    words = []   # extracted words
    dashes = []  # dash counts before/between/after words

    dash_count = 0
    i = 0

    # Parse into words and dash counts, ignoring spaces
    while i < len(text):
        c = text[i]
        if c == ' ':
            i += 1
        elif c == '-':
            dash_count += 1
            i += 1
        elif is_letter(c):
            # store dash count before this word
            dashes.append(dash_count)
            dash_count = 0

            # read the full word
            j = i
            while j < len(text) and is_letter(text[j]):
                j += 1
            words.append(text[i:j])
            i = j
        else:
            # any other character is invalid
            return "error"

    # dash count after last word
    dashes.append(dash_count)

    # If there are no words, the whole string is only spaces and hyphens
    if not words:
        if dash_count == 1:
            return "error"              # single hyphen cannot stand alone
        elif dash_count == 0:
            return '""'                 # empty after formatting
        else:
            return '"' + dash_string(dash_count) + '"'

    n = len(words)

    # Single hyphen cannot be at start or end (needs words on both sides)
    if dashes[0] == 1 or dashes[n] == 1:
        return "error"

    out_parts = []

    # leading dashes
    if dashes[0] > 0:
        out_parts.append(dash_string(dashes[0]))

    # first word
    out_parts.append(words[0])

    # between words
    for j in range(1, n):
        k = dashes[j]
        if k == 0:
            out_parts.append(" ")
        elif k == 1:
            out_parts.append("-")
        else:
            out_parts.append(dash_string(k))
        out_parts.append(words[j])

    # trailing dashes
    if dashes[n] > 0:
        out_parts.append(dash_string(dashes[n]))

    return '"' + "".join(out_parts) + '"'

def main():
    data = sys.stdin.read().splitlines()
    if not data:
        return
    t = int(data[0].strip())
    ans = []
    # Next t lines are the quoted texts (may be empty inside quotes, but line exists)
    for idx in range(1, 1 + t):
        ans.append(solve_one(data[idx] if idx < len(data) else ""))
    sys.stdout.write("\n".join(ans))

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

---

## 5) Compressed editorial

Parse each line (inside quotes), ignoring spaces, into a list of words and counts of consecutive `-` before/between/after words. Any non-letter/space/`-` → `error`. A single `-` is a hyphen and is only allowed between two words, so if the leading or trailing dash count is 1 → `error`; also if there are no words and total dashes is 1 → `error`. Rebuild output: for each gap with dash count `k`: print `" "` if `k=0`, `"-"` if `k=1`, else represent `k` as one `" --- "` if odd plus short dashes `" -- "` for the remainder (even). Wrap result in quotes. Linear time per line.
