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 (Clang 10.0.0) (judge verdict: Wrong Answer)

#include<stdio.h>
#include<stdlib.h>
#include<stdbool.h>
#include<string.h>
#include<math.h>
#include<ctype.h>

typedef long long ll;

ll gcd(ll a, ll b) { if (b == 0) return a; return gcd(b, a % b); }

ll lcm(ll a, ll b) { return a / gcd(a, b) * b; }

ll asc(const void* a, const void* b) { return *(ll*)a - *(ll*)b; }

ll dsc(const void* a, const void* b) { return *(ll*)b - *(ll*)a; }

ll n, k, a[200001], max = 0;

bool check(ll d) {
    ll count = 0;
    for (int i=0; i < n; i++) {
        count += a[i] / d;
    }
    return (count <= k)?1:0;
}

ll  binary(void) {
    ll r, l;
    r = 0, l = max+1;
    while (l - r > 1) {
        ll mid = (r + l) / 2;
        if (check(mid)) l = mid;
        else r = mid;
    }
    return l;
}

int main() {
    scanf("%lld%lld", &n, &k);
    for (int i = 0; i < n; i++) {
        scanf("%lld", &a[i]);
        max = (max > a[i]) ? max : a[i];
    }
    ll ans = binary();
    printf("%lld", ans);

    return 0;
}