//假设一个数x刚好是可以满足条件的最小值，那就回到了01模型,依照题目案例
1....6...16
00000111111   就可以用二分来解决
#include <iostream>

using namespace std;
const int N = 100010;
typedef long long LL;
int val[N];
int n , m;
int num;

bool check(int mid){
    int temp = 0;
    int cnt = 0;
    for(int i=1;i<=n;i++){
        temp += val[i];
        if(temp > mid) return false;
        if(i+1 <= n && temp+val[i+1] <= mid) continue;
        cnt++;
        temp = 0;
    }


    return m >= cnt;
}


int find(){
    int l = 0 , r = num;
    int ans = 0;
    while(l <= r){
        int mid=(l+r)/2;
        if(check(mid)){
            ans = mid;
            r = mid-1;
        }else{
            l = mid+1;
        }
    }
    return ans;
}


int main(){
    scanf("%d%d",&n,&m);

    for(int i=1;i<=n;i++){
        scanf("%d",val+i);
        num += val[i];
    } 

    cout<<find()<<endl;

    return 0;
}