## 1. Abridged Problem Statement

Diputs notation represents non-negative integers using 9 digit characters, from least to most significant:

```text
_ . , - ~ = ' ^ "
```

Each digit may appear only a limited number of times:

```text
_ : 2
. : 3
, : 5
- : 7
~ : 11
= : 13
' : 17
^ : 19
" : 23
```

A Diputs number is written from most significant digit to least significant digit, with repeated characters grouped together. Zero is represented by `O`.

The input is arbitrary text containing:

- decimal numbers,
- Diputs numbers,
- other characters.

You must replace every decimal number with its Diputs representation, and every Diputs number with its decimal representation.

Parsing must be greedy: always take the longest valid decimal or Diputs number possible. Decimal numbers cannot have leading zeroes, so `020` is parsed as `0` and `20`.

If the total number of parsed numbers is odd, sort the numeric values increasingly before output, while preserving the original positions’ output notation types.

Input and output sizes are at most `10^6` bytes.

---

## 2. Detailed Editorial

### Diputs as a mixed-radix numeral system

The Diputs digits are:

```text
index:  0 1 2 3 4 5 6 7 8
char:   _ . , - ~ = ' ^ "
```

The maximum count of each digit is:

```text
2, 3, 5, 7, 11, 13, 17, 19, 23
```

For digit `i`, the number of possible counts is:

```text
max_count[i] + 1
```

For example, `_` can appear `0`, `1`, or `2` times, so it has base `3`.

The ordering rule says that numbers are ordered first by the count of the most significant digit `"`, then by `^`, then by `'`, and so on.

This is equivalent to a mixed-radix system.

Define:

```text
product[0] = 1
product[i + 1] = product[i] * (max_count[i] + 1)
```

Then the decimal value of a Diputs number is:

```text
value = count[0] * product[0]
      + count[1] * product[1]
      + ...
      + count[8] * product[8]
```

The maximum representable number is:

```text
product[9] - 1
```

which equals:

```text
836075519
```

Zero is special and is represented by `O`.

---

### Converting decimal to Diputs

Given a decimal number `n`.

If `n == 0`, output:

```text
O
```

Otherwise, compute each digit count:

```text
count[i] = (n / product[i]) % (max_count[i] + 1)
```

Then output the digits from most significant to least significant:

```text
" repeated count[8] times
^ repeated count[7] times
...
_ repeated count[0] times
```

---

### Converting Diputs to decimal

For a Diputs token, every character directly contributes the value of its digit.

For example:

```text
,..__
```

means:

```text
1 comma + 2 dots + 2 underscores
```

So its decimal value is:

```text
product[2] + 2 * product[1] + 2 * product[0]
```

The implementation simply iterates through the token and adds:

```text
product[index_of_character]
```

for each character.

---

### Greedy parsing

The input is arbitrary text. We scan from left to right.

At each position:

#### 1. Decimal number

If the current character is a digit:

- If it is `0`, parse exactly one character as number `0`.
- Otherwise, keep taking digits while the resulting decimal number does not exceed the maximum Diputs value.

This handles cases like:

```text
020
```

as:

```text
0 and 20
```

and very large digit strings are split greedily.

---

#### 2. Diputs zero

If the current character is `O`, parse it as Diputs zero.

---

#### 3. Nonzero Diputs number

If the current character is one of:

```text
_ . , - ~ = ' ^ "
```

parse the longest valid Diputs representation.

A valid Diputs representation must:

1. Be written from most significant to least significant.
2. Not exceed the maximum count for any digit.

Using digit indices from least significant to most significant, this means indices must be non-increasing while reading left to right.

For example:

```text
,..__
```

has indices:

```text
2, 1, 1, 0, 0
```

which is valid.

But:

```text
_.
```

has indices:

```text
0, 1
```

which is invalid as one Diputs number, so it is parsed as:

```text
_
.
```

Also:

```text
___
```

cannot be one number because `_` may appear at most twice, so it is parsed as:

```text
__
_
```

---

#### 4. Other characters

Any other character is copied unchanged and acts as a separator.

---

### Sorting rule

After tokenization, suppose we found `k` numbers.

Each token stores:

- its position in the input,
- its original length,
- its numeric value,
- whether it was originally decimal.

Normally, each token is converted to the opposite notation independently.

If `k` is odd:

1. Extract all token values.
2. Sort them.
3. Assign sorted values back to the token positions in order.

Important: the notation type of each position is preserved.

That is, if a position originally contained a decimal number, then after sorting it still outputs Diputs notation. If a position originally contained a Diputs number, then after sorting it still outputs decimal notation.

---

### Complexity

Let `N` be the input size.

Tokenization is linear:

```text
O(N)
```

Sorting happens only on the parsed numbers:

```text
O(K log K)
```

where `K <= N`.

Memory usage is:

```text
O(N)
```

---

## 3. Provided C++ Solution with Detailed Comments

```cpp
#include <bits/stdc++.h> // Includes almost all standard C++ headers.

using namespace std;

// Output operator for pairs.
// Not essential for this problem, but part of the author's template.
template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second;
}

// Input operator for pairs.
// Also unused in the main solution.
template<typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& x) {
    return in >> x.first >> x.second;
}

// Input operator for vectors.
// Reads every element of the vector.
template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    for(auto& x: a) {
        in >> x;
    }
    return in;
};

// Output operator for vectors.
// Prints elements separated by spaces.
template<typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
    for(auto x: a) {
        out << x << ' ';
    }
    return out;
};

// Stores the entire input text.
string input_text;

// Reads all standard input into input_text.
void read() {
    stringstream buffer;          // Temporary buffer.
    buffer << cin.rdbuf();        // Copy all of cin into the buffer.
    input_text = buffer.str();    // Store it as a string.
}

void solve() {
    // Diputs digits from least significant to most significant.
    // Note that the final character is the double quote.
    const string diputs_chars = "_.,-~='^\"";

    // Maximum allowed count for each Diputs digit.
    const int max_counts[9] = {2, 3, 5, 7, 11, 13, 17, 19, 23};

    // products[i] is the place value of Diputs digit i.
    // products[0] = 1 for '_'.
    // products[1] = 3 for '.', because '_' has 3 possible counts: 0, 1, 2.
    int64_t products[10];
    products[0] = 1;

    // Build mixed-radix place values.
    for(int i = 0; i < 9; i++) {
        products[i + 1] = products[i] * (max_counts[i] + 1);
    }

    // Maximum representable Diputs number.
    int64_t max_num = products[9] - 1;

    // sig_of[c] gives the Diputs digit index of character c.
    // If c is not a Diputs digit, sig_of[c] is -1.
    array<int, 256> sig_of;
    sig_of.fill(-1);

    // Fill lookup table for Diputs characters.
    for(int i = 0; i < 9; i++) {
        sig_of[(unsigned char)diputs_chars[i]] = i;
    }

    // Lambda converting a substring input_text[start:end] from Diputs to decimal.
    auto diputs_to_decimal = [&](size_t start, size_t end) -> int64_t {
        int64_t value = 0;

        // Each Diputs character contributes its mixed-radix place value.
        for(size_t k = start; k < end; k++) {
            value += products[sig_of[(unsigned char)input_text[k]]];
        }

        return value;
    };

    // Lambda converting decimal n to Diputs notation.
    auto decimal_to_diputs = [&](int64_t n) -> string {
        // Zero is represented specially.
        if(n == 0) {
            return "O";
        }

        // counts[i] will be the number of occurrences of digit i.
        int counts[9];

        // Extract mixed-radix digits.
        for(int i = 0; i < 9; i++) {
            counts[i] = (n / products[i]) % (max_counts[i] + 1);
        }

        string result;

        // Output from most significant digit to least significant digit.
        for(int i = 8; i >= 0; i--) {
            result.append(counts[i], diputs_chars[i]);
        }

        return result;
    };

    // Represents one parsed number token.
    struct token {
        size_t start;       // Starting position in input_text.
        size_t length;      // Length of original token.
        int64_t value;      // Numeric value of the token.
        bool was_decimal;   // True if original token was decimal.
    };

    // All parsed numeric tokens.
    vector<token> tokens;

    // Length of input.
    size_t n = input_text.size();

    // Scan the input greedily from left to right.
    for(size_t i = 0; i < n;) {
        // Current character.
        unsigned char c = input_text[i];

        // Case 1: decimal number.
        if(isdigit(c)) {
            size_t j = i;       // End position of token.
            int64_t value = 0;  // Parsed decimal value.

            // Decimal numbers cannot have leading zeroes.
            // Therefore a token beginning with '0' has length exactly 1.
            if(c == '0') {
                j = i + 1;
            } else {
                // Greedily extend the decimal number while:
                // - the next character is a digit,
                // - the value does not exceed max_num.
                while(j < n && isdigit((unsigned char)input_text[j])) {
                    int64_t next_val = value * 10 + (input_text[j] - '0');

                    // If adding this digit would exceed the maximum,
                    // stop before it. The next scan iteration will start there.
                    if(next_val > max_num) {
                        break;
                    }

                    value = next_val;
                    j++;
                }
            }

            // Store this decimal token.
            tokens.push_back({i, j - i, value, true});

            // Continue scanning after the token.
            i = j;
        }

        // Case 2: Diputs zero, represented by capital letter O.
        else if(c == 'O') {
            tokens.push_back({i, 1, 0, false});
            i++;
        }

        // Case 3: nonzero Diputs number.
        else if(sig_of[c] >= 0) {
            size_t j = i;       // End position of token.
            int last_sig = 8;   // Previous digit index. Start with highest allowed.
            int counts[9] = {}; // Counts of each digit in this token.

            // Greedily extend while the representation stays valid.
            while(j < n) {
                // Digit index of current character, or -1 if not Diputs.
                int s = sig_of[(unsigned char)input_text[j]];

                // Stop if:
                // - this is not a Diputs digit,
                // - significance increases, violating most-to-least order,
                // - this digit already reached its maximum count.
                if(s < 0 || s > last_sig || counts[s] >= max_counts[s]) {
                    break;
                }

                // Use this character.
                counts[s]++;

                // Future digits must be no more significant than this one.
                last_sig = s;

                // Advance token end.
                j++;
            }

            // Store this Diputs token and compute its decimal value.
            tokens.push_back({i, j - i, diputs_to_decimal(i, j), false});

            // Continue scanning after the token.
            i = j;
        }

        // Case 4: ordinary text character.
        else {
            i++;
        }
    }

    // If the number of parsed numbers is odd, sort their numeric values.
    if(tokens.size() % 2 == 1) {
        vector<int64_t> values;
        values.reserve(tokens.size());

        // Extract values.
        for(auto& t: tokens) {
            values.push_back(t.value);
        }

        // Sort ascending.
        sort(values.begin(), values.end());

        // Put sorted values back into token positions.
        // The token notation type is not changed.
        for(size_t k = 0; k < tokens.size(); k++) {
            tokens[k].value = values[k];
        }
    }

    // Build final output.
    string result;

    // Reserve enough memory to avoid frequent reallocations.
    result.reserve(input_text.size() * 2);

    // cursor tracks how much of input_text has already been copied.
    size_t cursor = 0;

    // Process every parsed token in order.
    for(auto& t: tokens) {
        // Copy unchanged text before this token.
        while(cursor < t.start) {
            result += input_text[cursor++];
        }

        // If the original token was decimal, output Diputs.
        if(t.was_decimal) {
            result += decimal_to_diputs(t.value);
        }

        // If the original token was Diputs, output decimal.
        else {
            result += to_string(t.value);
        }

        // Skip the original token in the input.
        cursor = t.start + t.length;
    }

    // Copy remaining text after the last token.
    while(cursor < input_text.size()) {
        result += input_text[cursor++];
    }

    // Print final transformed text.
    cout << result;
}

int main() {
    // Fast I/O.
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

    // There is only one test case: the entire input.
    int T = 1;

    // Process the single test.
    for(int test = 1; test <= T; test++) {
        read();
        solve();
    }

    return 0;
}
```

---

## 4. Python Solution with Detailed Comments

```python
import sys


def main():
    # Read the entire input as one string.
    text = sys.stdin.read()

    # Diputs digits from least significant to most significant.
    diputs_chars = "_.,-~='^\""

    # Maximum allowed counts for each Diputs digit.
    max_counts = [2, 3, 5, 7, 11, 13, 17, 19, 23]

    # products[i] is the mixed-radix place value of digit i.
    products = [1]

    # Build place values.
    # products[i + 1] = products[i] * number_of_possible_counts_for_digit_i
    for cnt in max_counts:
        products.append(products[-1] * (cnt + 1))

    # Maximum representable value.
    max_num = products[9] - 1

    # Map each Diputs character to its significance index.
    sig_of = {ch: i for i, ch in enumerate(diputs_chars)}

    def diputs_to_decimal(s):
        """
        Convert a valid nonzero Diputs representation to decimal.
        """
        value = 0

        # Each character contributes its place value.
        for ch in s:
            value += products[sig_of[ch]]

        return value

    def decimal_to_diputs(n):
        """
        Convert a decimal integer to Diputs notation.
        """
        # Zero is special.
        if n == 0:
            return "O"

        counts = [0] * 9

        # Extract mixed-radix digits.
        for i in range(9):
            counts[i] = (n // products[i]) % (max_counts[i] + 1)

        result = []

        # Output from most significant digit to least significant digit.
        for i in range(8, -1, -1):
            if counts[i]:
                result.append(diputs_chars[i] * counts[i])

        return "".join(result)

    # Each token is represented as:
    # [start_position, length, numeric_value, was_decimal]
    tokens = []

    n = len(text)
    i = 0

    # Greedily scan input from left to right.
    while i < n:
        c = text[i]

        # Case 1: decimal number.
        if c.isdigit():
            j = i
            value = 0

            # Decimal numbers may not have leading zeroes.
            # Therefore '0' is always a token of length 1.
            if c == "0":
                j = i + 1
            else:
                # Extend while the token remains a valid decimal number
                # not exceeding max_num.
                while j < n and text[j].isdigit():
                    next_value = value * 10 + (ord(text[j]) - ord("0"))

                    # Stop before exceeding the representable range.
                    if next_value > max_num:
                        break

                    value = next_value
                    j += 1

            # Store decimal token.
            tokens.append([i, j - i, value, True])

            # Continue after token.
            i = j

        # Case 2: Diputs zero.
        elif c == "O":
            tokens.append([i, 1, 0, False])
            i += 1

        # Case 3: nonzero Diputs number.
        elif c in sig_of:
            j = i

            # Previous significance index.
            # A valid Diputs representation must have non-increasing
            # significance from left to right.
            last_sig = 8

            # Counts used in this token.
            counts = [0] * 9

            # Greedily extend the Diputs token.
            while j < n:
                ch = text[j]

                # Stop if not a Diputs digit.
                if ch not in sig_of:
                    break

                s = sig_of[ch]

                # Stop if significance increases.
                if s > last_sig:
                    break

                # Stop if this digit has reached its maximum allowed count.
                if counts[s] >= max_counts[s]:
                    break

                # Accept this character.
                counts[s] += 1

                # Future characters must be no more significant.
                last_sig = s

                # Advance end position.
                j += 1

            # Convert the parsed substring to decimal.
            value = diputs_to_decimal(text[i:j])

            # Store Diputs token.
            tokens.append([i, j - i, value, False])

            # Continue after token.
            i = j

        # Case 4: ordinary non-number text.
        else:
            i += 1

    # If there is an odd number of numeric tokens, sort their values.
    if len(tokens) % 2 == 1:
        values = sorted(token[2] for token in tokens)

        # Put sorted values back into token positions.
        # The original notation type at each position is preserved.
        for idx, value in enumerate(values):
            tokens[idx][2] = value

    # Build the final output.
    result = []
    cursor = 0

    for start, length, value, was_decimal in tokens:
        # Copy unchanged text before this token.
        if cursor < start:
            result.append(text[cursor:start])

        # Convert token to the opposite notation.
        if was_decimal:
            result.append(decimal_to_diputs(value))
        else:
            result.append(str(value))

        # Skip original token.
        cursor = start + length

    # Copy remaining text after the last token.
    if cursor < n:
        result.append(text[cursor:])

    # Write final answer.
    sys.stdout.write("".join(result))


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

---

## 5. Compressed Editorial

Treat Diputs notation as a mixed-radix system.

For digit counts:

```text
max = [2, 3, 5, 7, 11, 13, 17, 19, 23]
```

define:

```text
prod[0] = 1
prod[i + 1] = prod[i] * (max[i] + 1)
```

Then a Diputs number has value:

```text
sum(count[i] * prod[i])
```

and decimal-to-Diputs is done by:

```text
count[i] = (n // prod[i]) % (max[i] + 1)
```

Zero is represented by `O`.

Scan the entire input greedily:

- If current character is a decimal digit:
  - parse `0` as a single digit,
  - otherwise consume digits while value does not exceed the maximum representable number.
- If current character is `O`, parse Diputs zero.
- If current character is a Diputs digit, consume the longest valid Diputs token:
  - digit significances must be non-increasing,
  - no digit count may exceed its maximum.
- Otherwise skip it as normal text.

Store each token’s position, length, numeric value, and original notation type.

If the number of tokens is odd, sort only their numeric values and assign them back to token positions. The notation type of each position stays unchanged.

Finally reconstruct the text, copying separators unchanged and converting each token to the opposite notation.

Complexity:

```text
O(N + K log K)
```

where `N` is input size and `K` is the number of parsed numbers.