# GeetCode Hub

In a group of N people (labelled `0, 1, 2, ..., N-1`), each person has different amounts of money, and different levels of quietness.

For convenience, we'll call the person with label `x`, simply "person `x`".

We'll say that `richer[i] = [x, y]` if person `x` definitely has more money than person `y`.  Note that `richer` may only be a subset of valid observations.

Also, we'll say `quiet[x] = q` if person x has quietness `q`.

Now, return `answer`, where `answer[x] = y` if `y` is the least quiet person (that is, the person `y` with the smallest value of `quiet[y]`), among all people who definitely have equal to or more money than person `x`.

Example 1:

```Input: richer = [[1,0],[2,1],[3,1],[3,7],[4,3],[5,3],[6,3]], quiet = [3,2,5,4,6,1,7,0]
Output: [5,5,2,5,4,5,6,7]
Explanation:
Person 5 has more money than 3, which has more money than 1, which has more money than 0.
The only person who is quieter (has lower quiet[x]) is person 7, but
it isn't clear if they have more money than person 0.

Among all people that definitely have equal to or more money than person 7
(which could be persons 3, 4, 5, 6, or 7), the person who is the quietest (has lower quiet[x])
is person 7.

The other answers can be filled out with similar reasoning.
```

Note:

1. `1 <= quiet.length = N <= 500`
2. `0 <= quiet[i] < N`, all `quiet[i]` are different.
3. `0 <= richer.length <= N * (N-1) / 2`
4. `0 <= richer[i][j] < N`
5. `richer[i][0] != richer[i][1]`
6. `richer[i]`'s are all different.
7. The observations in `richer` are all logically consistent.

class Solution { public int[] loudAndRich(int[][] richer, int[] quiet) { } }