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 Java (OpenJDK 11.0.6) (judge verdict: Accepted)

import java.io.*;
import java.util.*;
import static java.lang.System.out;

public class Main{
    static MyReader in = new MyReader();

    public static void main(String[] args){
        int N = in.i();
        int K = in.i();
        int[] A = in.ii(N);
        int l = 1;
        int r = 1000000000;
        while (l < r) {
            int m = (l + r) / 2;
            int k = 0;
            for (int i = 0; i < A.length; i++) {
                k += (A[i] - 1) / m;
            }
            if (k <= K) {
                r = m;
            } else {
                l = m + 1;
            }
        }
        out.println(r);
    }

    static class MyReader extends BufferedReader {
        char[] cbuf = new char[1024];
        int head = 0;
        int tail = 0;

        MyReader() {
            super(new InputStreamReader(System.in));
        }

        char next() {
            if (head == tail) {
                try {
                    tail = read(cbuf, 0, cbuf.length);
                } catch (IOException e) {}
                head = 0;
            }
            return cbuf[head++];
        }

        void back() {
            head--;
        }

        boolean minus() {
            boolean minus;
            while (true) {
                char c = next();
                if (c != ' ' && c != '\n' && c != '\r') {
                    if (!(minus = c == '-')) back();
                    return minus;
                }
            }
        }

        void skip() {
            while (true) {
                char c = next();
                if (c != ' ' && c != '\n' && c != '\r') {
                    back();
                    break;
                }
            }
        }

        char[] s(final int N) {
            skip();
            char[] s = new char[N];
            for (int i = 0; i < s.length; i++) {
                s[i] = next();
            }
            return s;
        }

        int i() {
            boolean minus = minus();
            int n = 0;
            while (true) {
                int k = next() - '0';
                if (k < 0 || 9 < k) break;
                n = 10 * n + k;
            }
            return minus ? -n : n;
        }

        int[] ii(final int N) {
            int[] a = new int[N];
            for (int j = 0; j < a.length; j++) a[j] = i();
            return a;
        }

        long l() {
            boolean minus = minus();
            long n = 0;
            while (true) {
                int k = next() - '0';
                if (k < 0 || 9 < k) break;
                n = 10 * n + k;
            }
            return minus ? -n : n;
        }
    }
}