/*
　　题意：给定n,m，然后n个数字，要求一个最小的lim
　　使得这连续n个数字可以被分为连续的m个集合，每个集合的和都不大于Lim
/*
#include <iostream>#include <cstdio>#define range(i,a,b) for (int i=a;i<=b;i++)using namespace std;const int maxn =100000;int Cost[maxn+1];int L,R;int n,m;bool check(int val){    int sum(0);    int ans(0);    range(i,1,n)        if (Cost[i]>val)            return 0;        else        if (sum+Cost[i]<=val)        {            sum+=Cost[i];        }        else        {            sum=Cost[i];            ans++;        }    if (sum!=0)        ans++;    return ans<=m;}int main(){    cin>>n>>m;    L = R = 0;    range(i,1,n)    {        scanf("%d",&Cost[i]);        R+=Cost[i];    }    range(c,1,100)    {        int M = (L+R)>>1;        if (check(M))            R=M;        else            L=M;    }    if (check(L))        cout<<L<<endl;    else        cout<<R<<endl;    return 0;}