Given a list of elements, find the frequency of elements that are greater than or equal to k.

For example, a list of `[3, 2, 3, 1, 1, 1]` and k=2 should return `[1, 3]`.

```javascript
function top_k(elements: number[], k: number) {
  const set: { [key: number]: number } = {}

for (let e of elements) {
    const count = set[e] ?? 0
    set[e] = count + 1
  }

let output: number[] = []

const entries: [string, number][] = Object.entries(set)
  entries.sort((a, b) => b[1] - a[1])

let count = 0

for (let [key, v] of entries) {
    if (count < k) {
      output.push(Number(key))
      count++
    } else {
      break
    }
  }

console.log(output)
}

// { 3: 2, 2: 1, 1: 3}
top_k([3, 2, 3, 1, 1, 1], 2)
top_k([3, 2, 3, 1, 1, 1], 3)
top_k([3, 2, 3, 1, 1, 1], 1)
```