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: Accepted)

#include <float.h>
#include <inttypes.h>
#include <limits.h>
#include <signal.h>
#include <stdarg.h>
#include <stdbool.h>
#include <stdint.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <time.h>
#include <unistd.h>
#include <math.h>
#ifdef __cplusplus
#include <bits/stdc++.h>
#endif
//#pragma GCC target("avx2")
//#pragma GCC optimize("O3")
//#pragma GCC optimize("unroll-loops")

#define getchar getchar_unlocked
#define putchar putchar_unlocked
int32_t nextint(void){ char c=getchar(); while(c!='-'&&(c<'0'||'9'<c)) c=getchar(); bool s=false; if(c=='-'){s=true;c=getchar();} uint32_t x=0; while('0'<=c && c<='9'){ x=x*10+c-'0'; c=getchar(); } return s?-x:x; }
int64_t nextlong(void){ char c=getchar(); while(c!='-'&&(c<'0'||'9'<c)) c=getchar(); int s=0; if(c=='-'){s=1;c=getchar();} uint64_t x=0; while('0'<=c && c<='9'){ x=x*10+c-'0'; c=getchar(); } return s?-x:x; }
uint32_t nextstr(char *s){ char c=getchar(); while(c==' '||c=='\n') c=getchar(); uint32_t len=0; while(c!=' '&&c!='\n'){ *s++=c; len++; c=getchar(); } *s='\0'; return len; }
void rsorta_d32(void *base, const int count, const int size, const int index){
	typedef int32_t rsort_t;
	rsort_t (*a)[size] = (rsort_t(*)[size])base;
	rsort_t (*b)[size] = (rsort_t(*)[size])malloc(count*size*sizeof(rsort_t));
	int c[0x100];
	for(int k=0; k<4; k++){
		memset(c, 0, sizeof(c)); for(int i=0; i<count; i++){ c[(a[i][index]>>(k<<3))&0xFF]++; } for(int i=1; i<0x100; i++) c[i]+=c[i-1]; for(int i=count-1; i>=0; i--){ int to=--c[(a[i][index]>>(k<<3))&0xFF]; for(int j=0; j<size; j++){ b[to][j]=a[i][j]; } }{rsort_t(*t)[size]=a; a=b; b=t;}
	}
	free(b);
}

int n;
int a[200000];

inline int f(int k){
	long ans=0;
	for(int i=0; i<n; i++){
		ans+=(a[i]-1)/k;
	}
	return ans>INT_MAX?INT_MAX:ans;
}

int main(void){
	n=nextint();
	int k=nextint();
	for(int i=0; i<n; i++){
		a[i]=nextint();
	}
	int l=1, r=1000000000;
	while(l<=r){
		int m=(l+r)/2;
		if(f(m)<=k) r=m-1; else l=m+1;
	}
	printf("%d\n", l);
}