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 ≤ moneyi ≤ 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

7 5
100
400
300
100
500
101
400

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.

给出农夫在n天中每天的花费，要求把这n天分成连续的m组，要求各组的花费之和应该尽可能地小，最后输出各组花费之和中的最大值。

//#include<bits/stdc++.h>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define int long long
#define N 1000005
#define INF 0x3f3f3f3f
using namespace std;
int a[N];
int n,m;
int r,l;
int ans,mid;
bool check(int k)
{
    int sum=0,num=0;
    for(int i=1;i<=n;i++)
    {
        if(sum+a[i]<=k) sum+=a[i]; //给每个月的额定值开始分配，如果还没装满的话，那就接着装，
        
        else {    //如果再装要溢出的话那就换一个新的装
            sum=a[i];
            num++;  //记录装完所需的组数
        }
    }
    return num>=m;  //比较装完后的组数是否合规
}
signed main()
 {
    ios::sync_with_stdio(false);
  cin>>n>>m;
  for(int i=1;i<=n;i++)
  {cin>>a[i];
   l=max(l,a[i]); //计算一个月最差情况所需的最小花费
   r+=a[i]; //预处理，计算所有月份所花费之和，也就是右边界
  }
 
  while(l<=r)
  {
     mid=l+r>>1;   
    if(check(mid)) l=mid+1;
    else r=mid-1;
    ans=l;
  }
cout<<ans<<endl;
    return 0; 
 }