Count anagram-based replacements per phrase

## Problem You are given: - `phrases`: a list of strings, where each string is a phrase containing words separated by single spaces. - `words`: a list of strings. A word `a` can be **replaced** by a word `b` if `a` and `b` are **anagrams** (they contain the same letters with the same frequencies; order does not matter). For each phrase, compute how many different phrases can be formed by replacing **each replaceable word** in the phrase with **any** anagram from `words`. - Words in the phrase that have **no anagram in `words`** must stay unchanged. - If **no word** in the phrase has any anagram in `words` (i.e., no replacement is possible), return `0` for that phrase. Return a list of integers, one per phrase. ### Example Input: - `phrases = ["hello world", "below elbow"]` - `words = ["below", "elbow"]` Explanation: - Phrase 1: neither `hello` nor `world` has an anagram in `words` → no replacement possible → `0` - Phrase 2: `below` and `elbow` are anagrams, and both appear in `words`. Each of the two positions can be replaced by either `below` or `elbow` → `2 * 2 = 4` Output: - `[0, 4]` ## Requirements / Notes - Treat anagrams case-sensitively or case-insensitively consistently (state your choice in implementation; a common assumption is lowercase inputs). - You may assume phrases contain only alphabetic words and spaces. - The result may exceed 32-bit integer range; use 64-bit if needed.