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 (GCC 9.2.1) (judge verdict: Runtime Error)

#define _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <string.h>
#include <stdbool.h>
typedef long long LL;
typedef long double Lf;
#define in (void)scanf
#define Rep(i,a,n)for(int i=(int)(a);i<(int)(n);i++)
//#define Rep(i,a,n)for(LL i=(LL)(a);i<(LL)(n);i++)
#define rep(i,n)Rep(i,0,n)
#define Repp(i,l,r,k)for(int i=(int)(l);i<(int)(r);i+=(int)(k))
//#define Repp(i,l,r,k)for(LL i=(LL)(l);i<(LL)(r);i+=(LL)(k))
#define Ma(x,y) ((x)>(y)?(x):(y))
#define mi(x,y) ((x)<(y)?(x):(y))
#define MOD 1000000007//998244353,1000000007,1000000009,4999999999999997
int upi(const void* a, const void* b) { return*(int*)a<*(int*)b ? -1 : *(int*)a> * (int*)b ? 1 : 0; }
int downi(const void* a, const void* b) { return*(int*)a<*(int*)b ? 1 : *(int*)a> * (int*)b ? -1 : 0; }
void sortupi(int* a, int n) { qsort(a, n, sizeof(int), upi); } //1,2,3,4,...
void sortdowni(int* a, int n) { qsort(a, n, sizeof(int), downi); }
int upLL(const void* a, const void* b) { return*(LL*)a<*(LL*)b ? -1 : *(LL*)a> * (LL*)b ? 1 : 0; }
int downLL(const void* a, const void* b) { return*(LL*)a<*(LL*)b ? 1 : *(LL*)a> * (LL*)b ? -1 : 0; }
void sortupLL(LL* a, int n) { qsort(a, n, sizeof(LL), upLL); }
void sortdownLL(LL* a, int n) { qsort(a, n, sizeof(LL), downLL); }
int cmp_char(const void* a, const void* b) { return *(char*)a - *(char*)b; } // a,b,c,d
int cmp_str(const void* a, const void* b) { return strcmp(*(const char**)a, *(const char**)b); } // aaa,aab
LL gcd(LL p, LL q) { while (q) { LL t = p % q; p = q; q = t; }return p; }
LL modpow(LL a,LL n,int m){LL x=1;for(a%=m;n;n/=2)n&1?x=x*a%m:0,a=a*a%m;return x;}
#define invmodpow(b,p)modpow(b,p-2,p)// a/b≡a*b^{-1}(mod p)...return b^{-1}(mod p)
typedef struct fac { LL n, exp; }F;
LL fact(LL n,F*a){LL x=n,i,j,c=0;for(i=2;i*i<=x;i++){for(j=1;x%i==0;j++)x/=i;if(j>1){a[c].n=i;a[c++].exp=j-1;}}if(x-1){a[c].n=x;a[c++].exp=1;}return c;}
//snippet

#define N 100009

int n, k, a[N];

LL check(int x) {
	LL ret = 0;
	rep(i, n)ret += (LL)ceill((Lf)a[i] / (Lf)x) - 1;
	return ret;
}

void run(void) {
	in("%d%d", &n, &k);
	rep(i, n)in("%d", &a[i]);

	int p = 0, q = MOD / 2, r = MOD, ans = n;
	while (r - p > 1) {
		if (check(q) <= (LL)k) r = q;
		else p = q;
		q = (p + r) / 2;
	}
	if (check(r) <= k)ans = r;
	if (check(p) <= k)ans = p;
	printf("%d\n", ans);
}

int main(void) {
	int test = 1;
//	in("%d", &test);
	while (test--) {
		run();
	}
	return 0;
}