## 1) Abridged problem statement

You're given a syntactically correct boolean expression over variables `a..j` with operators:

- unary: `!` (NOT)
- binary: `&` (AND), `||` (OR), `<=>` (equivalence), `=>` (implication), `#` (XOR)

Operator precedence: `!` highest, `&` middle, all others lowest. Parentheses can appear.

Output an **equivalent** expression (same truth table for all assignments of `a..j`) that contains **no parentheses**, and whose length is ≤ 32768.

---

## 2) Detailed editorial (solution idea)

### Key observations
1. There are only **10 variables** (`a..j`), so there are at most `2^10 = 1024` different assignments of truth values.
2. If we can evaluate the given expression for every assignment, we know exactly which assignments make it `true`.
3. From those satisfying assignments we can build a **DNF** (Disjunctive Normal Form):
   - For each assignment that makes the expression true, create a conjunction (AND-chain) describing that assignment:
     - variable is `x` if it's true
     - variable is `!x` if it's false
   - OR (`||`) together all such conjunctions.
4. Such a DNF needs **no parentheses** because `&` has higher precedence than `||`, so
   ```
   a&b||a&c
   ```
   is parsed as `(a&b) || (a&c)` automatically.

### Parsing / evaluating the original expression
We must correctly respect precedence and parentheses. A standard recursive-descent parser works well:

- `parse_low` handles the *lowest-precedence* binary operators: `||`, `<=>`, `=>`, `#`
- `parse_mid` handles `&`
- `parse_high` handles unary `!`
- `parse_atom` handles variables `a..j` and parenthesized subexpressions `( ... )`

Each parser function consumes characters from the global string using an index `pos` and returns a boolean value for a given assignment `mask` (10-bit mask).

### Building the DNF
1. Detect which variables actually appear in the expression (at most 10, often fewer). Let there be `k` used variables.
2. For each full 10-variable mask `0..1023`:
   - evaluate the expression
   - if it is true, compress the assignment down to only the `k` used vars (so output clauses don't mention unused variables)
3. If there are **no** satisfying assignments, output a contradiction like `a&!a`.
4. Otherwise, output all satisfying minterms:
   ```
   lit1 & lit2 & ... & litk || ... || ...
   ```
   without parentheses; join literals in a clause with `&` and clauses with `||`.

### Correctness sketch
- The parser evaluates the expression exactly according to the problem's grammar and precedence rules.
- The produced DNF is true on exactly the same assignments as the original formula (by construction, it ORs precisely the satisfying assignments).
- Therefore, the output is logically equivalent and parenthesis-free.

### Complexity
- Evaluate expression up to 1024 times.
- Each evaluation scans the expression (≤ 2048 chars) with the parser.
- Upper bound is easily fast enough: about ~2 million character operations. The final string is bounded by 2^vars · (vars·3 + 1) characters, comfortably under 32K.

---

## 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 expr;

void read() { cin >> expr; }

int pos;

bool parse_low(int mask);
bool parse_mid(int mask);
bool parse_high(int mask);
bool parse_atom(int mask);

bool parse_atom(int mask) {
    if(expr[pos] == '(') {
        pos++;
        bool result = parse_low(mask);
        pos++;
        return result;
    }
    char var = expr[pos++];
    return (mask >> (var - 'a')) & 1;
}

bool parse_high(int mask) {
    if(expr[pos] == '!') {
        pos++;
        return !parse_high(mask);
    }
    return parse_atom(mask);
}

bool parse_mid(int mask) {
    bool result = parse_high(mask);
    while(pos < (int)expr.size() && expr[pos] == '&') {
        pos++;
        result = result & parse_high(mask);
    }
    return result;
}

bool parse_low(int mask) {
    bool result = parse_mid(mask);
    while(pos < (int)expr.size()) {
        if(expr[pos] == '|' && expr[pos + 1] == '|') {
            pos += 2;
            result = result | parse_mid(mask);
        } else if(expr[pos] == '<' && expr[pos + 1] == '=' && expr[pos + 2] == '>') {
            pos += 3;
            result = result == parse_mid(mask);
        } else if(expr[pos] == '=' && expr[pos + 1] == '>') {
            pos += 2;
            result = !result | parse_mid(mask);
        } else if(expr[pos] == '#') {
            pos++;
            result = result != parse_mid(mask);
        } else {
            break;
        }
    }
    return result;
}

bool evaluate_expression(int mask) {
    pos = 0;
    return parse_low(mask);
}

string create_dnf() {
    vector<bool> used_vars(10, false);
    for(char c: expr) {
        if(c >= 'a' && c <= 'j') {
            used_vars[c - 'a'] = true;
        }
    }

    vector<int> used_indices;
    for(int i = 0; i < 10; i++) {
        if(used_vars[i]) {
            used_indices.push_back(i);
        }
    }

    int num_used = used_indices.size();
    set<int> true_assignments;

    for(int mask = 0; mask < (1 << 10); mask++) {
        if(evaluate_expression(mask)) {
            int assignment = 0;
            for(int j = 0; j < num_used; j++) {
                if((mask >> used_indices[j]) & 1) {
                    assignment |= (1 << j);
                }
            }
            true_assignments.insert(assignment);
        }
    }

    if(true_assignments.empty()) {
        return "a&!a";
    }

    string result;
    bool first = true;
    for(int assignment: true_assignments) {
        if(!first) {
            result += "||";
        }
        first = false;
        for(int j = 0; j < num_used; j++) {
            if(j > 0) {
                result += '&';
            }
            if(!((assignment >> j) & 1)) {
                result += '!';
            }
            result += char('a' + used_indices[j]);
        }
    }
    return result;
}

void solve() {
    // Notice that there are 10 variables, meaning there are only 2^10 possible
    // results. This means that if we evaluate all of them, we could in theory
    // look at which values the output is true, and enumerate all of them as a
    // DNF (disjunctive normal form), without brackets, because the conjunction
    // has higher priority than disjunction. Not very relevant to the problem,
    // but if we were to have the opposite priority, one approach could have
    // been to go for a CNF, which can be done by having a DNF of !f(a,...,j),
    // and then negating the resulting DNF with De Morgan's Law.
    //
    // In terms of complexity, we could do this by naively evaluating, but a
    // slightly better approach is to recursively do it and return the full
    // 2^vars table as the constant would be a bit lower and we don't have to
    // worry a lot about how quickly we are searching for the lowest priority
    // operator. However, that's not actually needed and we can go for other
    // standard evaluation approaches by either recursively simulating a stack,
    // or directly using inverse polish notation.
    //
    // The final expression will have less than 2^vars * (vars*3 + 1)
    // characters, which is withing the bound of ~32K for 10 variables. The *3
    // is because we could in theory have a negation, the variable, and the &
    // operator between each atom in each clause. The original expression has
    // length less than 2048 characters, so even the naive approach of
    // evaluating each combination would have been fast enough.

    string result = create_dnf();
    cout << result << '\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 (same approach) with detailed comments

```python
import sys

# We implement the same recursive-descent parser as the C++ code.
# expr: input expression string
# pos: current index into expr while parsing

def solve():
    expr = sys.stdin.readline().strip()
    pos = 0  # parsing pointer

    def parse_low(mask: int) -> bool:
        """Parse lowest-precedence binary operators: ||, <=>, =>, #."""
        nonlocal pos
        result = parse_mid(mask)

        while pos < len(expr):
            # OR: "||"
            if pos + 1 < len(expr) and expr[pos] == '|' and expr[pos + 1] == '|':
                pos += 2
                result = result or parse_mid(mask)

            # Equivalence: "<=>"
            elif pos + 2 < len(expr) and expr[pos:pos+3] == "<=>":
                pos += 3
                result = (result == parse_mid(mask))

            # Implication: "=>", where (A => B) == (!A) or B
            elif pos + 1 < len(expr) and expr[pos:pos+2] == "=>":
                pos += 2
                result = ((not result) or parse_mid(mask))

            # XOR: "#"
            elif expr[pos] == '#':
                pos += 1
                result = (result != parse_mid(mask))

            else:
                break

        return result

    def parse_mid(mask: int) -> bool:
        """Parse conjunction '&' (middle precedence)."""
        nonlocal pos
        result = parse_high(mask)

        while pos < len(expr) and expr[pos] == '&':
            pos += 1
            result = result and parse_high(mask)

        return result

    def parse_high(mask: int) -> bool:
        """Parse unary '!' (highest precedence) or atom."""
        nonlocal pos
        if pos < len(expr) and expr[pos] == '!':
            pos += 1
            return not parse_high(mask)
        return parse_atom(mask)

    def parse_atom(mask: int) -> bool:
        """Parse a variable a..j or a parenthesized subexpression."""
        nonlocal pos

        # Parenthesized expression
        if expr[pos] == '(':
            pos += 1              # consume '('
            val = parse_low(mask) # parse inside
            pos += 1              # consume ')'
            return val

        # Variable
        ch = expr[pos]
        pos += 1
        bit = ord(ch) - ord('a')  # 0..9
        return ((mask >> bit) & 1) == 1

    def eval_expr(mask: int) -> bool:
        """Evaluate full expression under assignment mask."""
        nonlocal pos
        pos = 0
        return parse_low(mask)

    # Determine which variables appear (to avoid outputting unused ones).
    used = [False] * 10
    for c in expr:
        if 'a' <= c <= 'j':
            used[ord(c) - ord('a')] = True

    used_indices = [i for i in range(10) if used[i]]
    k = len(used_indices)

    true_assignments = set()

    # Enumerate all 10-variable assignments
    for mask in range(1 << 10):
        if eval_expr(mask):
            # Compress to just the used variables to keep output shorter
            a = 0
            for j, idx in enumerate(used_indices):
                if (mask >> idx) & 1:
                    a |= (1 << j)
            true_assignments.add(a)

    # If unsatisfiable: output a contradiction without parentheses.
    if not true_assignments:
        print("a&!a")
        return

    # Build DNF: OR over all satisfying minterms.
    clauses = []
    for a in sorted(true_assignments):
        lits = []
        for j, idx in enumerate(used_indices):
            var = chr(ord('a') + idx)
            if ((a >> j) & 1) == 0:
                lits.append('!' + var)
            else:
                lits.append(var)
        clauses.append("&".join(lits))

    print("||".join(clauses))


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

---

## 5) Compressed editorial

- With only variables `a..j`, there are at most `1024` assignments.
- Parse and evaluate the expression for every assignment (recursive descent with precedence: `!` > `&` > others).
- Collect all assignments where the expression is `true`.
- Output their DNF: for each satisfying assignment print a conjunction of literals (`x` or `!x`) over used variables, join literals by `&`, and join clauses by `||`.
- No parentheses are needed since `&` binds tighter than `||`.
- If no assignment satisfies the expression, print `a&!a`.
