Let the function `f(s)` be the frequency of the lexicographically smallest character in a non-empty string `s`. For example, if `s = "dcce"` then `f(s) = 2` because the lexicographically smallest character is `'c'`, which has a frequency of 2.

You are given an array of strings `words` and another array of query strings `queries`. For each query `queries[i]`, count the number of words in `words` such that `f(queries[i])` < `f(W)` for each `W` in `words`.

Return an integer array `answer`, where each `answer[i]` is the answer to the `ith` query.

Example 1:

    Input: queries = ["cbd"], words = ["zaaaz"]
    Output: [1]
    Explanation: On the first query we have f("cbd") = 1, f("zaaaz") = 3 so f("cbd") < f("zaaaz").

Example 2:

    Input: queries = ["bbb","cc"], words = ["a","aa","aaa","aaaa"]
    Output: [1,2]
    Explanation: On the first query only f("bbb") < f("aaaa"). On the second query both f("aaa") and f("aaaa") are both > f("cc").

Constraints:

- `1 <= queries.length <= 2000`
- `1 <= words.length <= 2000`
- `1 <= queries[i].length, words[i].length <= 10`
- `queries[i][j]`, `words[i][j]` consist of lowercase English letters.