#if 0 //内存爆，30%

#include<iostream>

#include<algorithm>

#include<cstring>

#include<cstdio>

using namespace std;

typedef long long ll;

ll a[], f[][];

int main()

{

    int n, m, i, j, k;

    ll maxt, tmp;

    while(scanf("%d%d",&n,&m)==)

    {

        memset(f,,sizeof(f));

        for(i=;i<=n;i++)

        {

            scanf("%d",&a[i]);

            f[i][] = f[i - ][] + a[i];

        }

        for (i = ; i <= m; i++)

        {

            for (j = i; j <= n; j++)

            {

                tmp = 2147483647L ; //tmp = LONG_MAX;

                for (k = i; k<j; k++)

                {

                    maxt = max(f[k][i - ], f[j][] - f[k][]);

                    if (tmp>maxt)

                        tmp = maxt;

                }

                f[j][i] = tmp;

            }

        }

        printf("%d\n", f[n][m]);

    }

    return ;

}

#endif

#if 0 //AC

#include <cstdio>

#include <algorithm>

using namespace std;

#define MAX(a,b) a>b?a:b   

typedef long long ll;

int n, m;

ll a[];

ll sum;

ll maxx;

bool judge(ll x)

{

    ll duan = ;//表示分的段数

    ll he = ;

    for (int i = ; i<n; i++)

    {

        if (he + a[i]>x)//当当前段再加一段长度后超过了x的值，那么新开一段

        {

            duan++;

            he = a[i];

            if (duan >= m)//当还没有扫完这个序列但是已有的段数已经超过了所规定的段数

            {

                return false;

            }

        }

        else

        {

            he = he + a[i];//延长该段

        }

    }

    return true;

}

int main()

{

    while (scanf("%d%d", &n, &m) == )

    {

        sum = ;

        maxx = -;

        for (int i = ; i < n; i++)

        {

            scanf("%lld", &a[i]);

            sum = sum + a[i];

            maxx = MAX(maxx, a[i]);

        }

        ll l = maxx, r = sum;//左区间用maxx是因为所分的段里面如果不含该序列的最大值，那么这个段子的和一定大于最大值,右区间为sum是因为不分的时候段子和为sum

        ll mid, ans;

        while (l <= r)

        {

            mid = (l + r) / ;

            if (judge(mid))//看是否能将序列分成符合段数不大于m 且最大和接近mid的段子们

            {

                ans = mid;

                r = mid - ;//求的最小值

            }

            else

            {

                l = mid + ;

            }

        }

        printf("%lld\n", ans);

    }

    return ;

}

#endif