Problem:
Score : 500 points

Problem Statement

We have N logs of lengths A_1,A_2,\cdots A_N.

We can cut these logs at most K times in total. When a log of length L is cut at a point whose distance from an end of the log is t (0<t<L), it becomes two logs of lengths t and L-t.

Find the shortest possible length of the longest log after at most K cuts, and print it after rounding up to an integer.

Constraints

1 \leq N \leq 2 \times 10^5

0 \leq K \leq 10^9

1 \leq A_i \leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N K
A_1 A_2 \cdots A_N

Output

Print an integer representing the answer.

Sample Input 1

2 3
7 9

Sample Output 1

4

First, we will cut the log of length 7 at a point whose distance from an end of the log is 3.5, resulting in two logs of length 3.5 each.

Next, we will cut the log of length 9 at a point whose distance from an end of the log is 3, resulting in two logs of length 3 and 6.

Lastly, we will cut the log of length 6 at a point whose distance from an end of the log is 3.3, resulting in two logs of length 3.3 and 2.7.

In this case, the longest length of a log will be 3.5, which is the shortest possible result. After rounding up to an integer, the output should be 4.

Sample Input 2

3 0
3 4 5

Sample Output 2

5

Sample Input 3

10 10
158260522 877914575 602436426 24979445 861648772 623690081 433933447 476190629 262703497 211047202

Sample Output 3

292638192

Solution in C# (Mono-mcs 6.8.0.105) (judge verdict: Time Limit Exceeded)

using System;
using System.Collections.Generic;
class Program
{
	static void Main(string[] args)
	{
		string[] str = Console.ReadLine().Split();
		int N = int.Parse(str[0]);
		int K = int.Parse(str[1]);
		int[]a = new int[N];
		int[]b = new int[N];
		string[] str2 = Console.ReadLine().Split();
		int[]KA = new int[N];
		var pq = new PriorityQueue<int>((x,y)=>(b[y].CompareTo(b[x])));
		for(var i=0;i<N;i++){
			a[i] = int.Parse(str2[i]);
			b[i] = a[i];
          	KA[i] = 1;
			pq.Enqueue(i);
		}
		for(var i=0;i<K;i++){
			int c = pq.Dequeue();
			KA[c]++;
			b[c] = (a[c]+KA[c]-1)/KA[c];
			pq.Enqueue(c);
		}
		int d = pq.Dequeue();
		Console.WriteLine((a[d]+KA[d]-1)/KA[d]);
	}
}

class PriorityQueue<T>
{
    private readonly List<T> heap;
    private readonly Comparison<T> compare;
    private int size;
    public PriorityQueue() : this(Comparer<T>.Default) {} 
    public PriorityQueue(IComparer<T> comparer) : this(16, comparer.Compare) { }
    public PriorityQueue(Comparison<T> comparison) : this(16, comparison) { }
    public PriorityQueue(int capacity, Comparison<T> comparison)
    {
        this.heap = new List<T>(capacity);
        this.compare = comparison;
    }
    public void Enqueue(T item)
    {
        this.heap.Add(item);
        var i = size++;
        while (i > 0)
        {
            var p = (i - 1) >> 1;
            if (compare(this.heap[p], item) <= 0)
                break;
            this.heap[i] = heap[p];
            i = p;
        }
        this.heap[i] = item;

    }
    public T Dequeue()
    {
        var ret = this.heap[0];
        var x = this.heap[--size];
        var i = 0;
        while ((i << 1) + 1 < size)
        {
            var a = (i << 1) + 1;
            var b = (i << 1) + 2;
            if (b < size && compare(heap[b], heap[a]) < 0) a = b;
            if (compare(heap[a], x) >= 0)
                break;
            heap[i] = heap[a];
            i = a;
        }
        heap[i] = x;
        heap.RemoveAt(size);
        return ret;
    }
    public T Peek() { return heap[0]; }
    public int Count { get { return size; } }
    public bool Any() { return size > 0; }
}