Farmer John is an astounding accounting wizard and has realized he might run out of money to run the farm. He has already calculated and recorded the exact amount of money (1 ≤ money_i ≤ 10,000) that he will need to spend each day over the next N (1 ≤ N ≤ 100,000) days.

FJ wants to create a budget for a sequential set of exactly M

(1 ≤ M ≤ N) fiscal periods called "fajomonths". Each of these fajomonths contains a set of 1 or more consecutive days. Every day is contained in exactly one fajomonth.

FJ's goal is to arrange the fajomonths so as to minimize the expenses of the fajomonth with the highest spending and thus determine his monthly spending limit.

input

Line 1: Two space-separated integers: N and M
Lines 2..N+1: Line i+1 contains the number of dollars Farmer John spends on the ith day

output

Line 1: The smallest possible monthly limit Farmer John can afford to live with.

sample input

7 5
100
400
300
100
500
101
400

sample output

500

Hint

If Farmer John schedules the months so that the first two days are a month, the third and fourth are a month, and the last three are their own months, he spends at most $500 in any month. Any other method of scheduling gives a larger minimum monthly limit. 題意：就是將序列劃分為k段，要求找出k段中最大的資金和要在所有符合要求的劃分中值最少  分析：比較典型的最大值最小化問題，用二分查詢可以很好的解決這個問題。這類問題的框架為，找出下界left和上界right， while(left< right)， 求出mid，看這個mid值是符合題意，
繼續二分。最後right即為答案。
本題中的下界為n個數中的最大值，因為這時候，是要劃分為n個區間（即一個數一個區間），left是滿足題意的n個區間和的最大值，
上屆為所有區間的和，因為這時候，是要劃分為1個區間（所有的數都在一個區間裡面）,    1<=m<=n， 所以我們所要求的值肯定在 [left, right] 之間。
對於每一個mid，遍歷一遍n個數，看能劃分為幾個區間，如果劃分的區間小於（或等於）給定的m，說明上界取大了， 那麼 另 right=mid，否則另 left=mid+1.

#include<cstdio>  
#include<cstring>  
#include<string>
#include<iostream>
#include<algorithm>
using namespace std;  
int a[100005];  
int main()  
{  
    int n,m,sum,maxn,i,j,s,cnt,mid;  
    while(~scanf("%d%d",&n,&m))  
    {  
        sum=0,maxn=0;  
        for(i=0;i<n;i++)  //首先求和和找到最大值。 
        {  
            scanf("%d",&a[i]);  
            sum+=a[i];  
            maxn=max(maxn,a[i]);  
        }  
        while(maxn<sum)  
        {  
            mid=(sum+maxn)/2;  
            s=0,cnt=0;  
            for(i=0;i<n;i++)  
            {  
                s+=a[i];  
                if(s>mid)  
                {  
                    s=a[i];  
                    cnt++;  
                }  
            }  
            if(cnt<m) 
			sum=mid;  
            else 
			maxn=mid+1;  
        }  
        printf("%d\n",maxn);  
    }    
}