Problem:
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You are given $n$ packages of $w_i$ kg from a belt conveyor in order ($i = 0, 1, ... n-1$). You should load all packages onto $k$ trucks which have the common maximum load $P$. Each truck can load consecutive packages (more than or equals to zero) from the belt conveyor unless the total weights of the packages in the sequence does not exceed the maximum load $P$.

Write a program which reads $n$, $k$ and $w_i$, and reports the minimum value of the maximum load $P$ to load all packages from the belt conveyor.

Input

In the first line, two integers $n$ and $k$ are given separated by a space character. In the following $n$ lines, $w_i$ are given respectively.

Output

Print the minimum value of $P$ in a line.

Constraints

$1 \leq n \leq 100,000$

$1 \leq k \leq 100,000$

$1 \leq w_i \leq 10,000$

Sample Input 1

5 3
8
1
7
3
9

Sample Output 1

10

If the first truck loads two packages of $\{8, 1\}$, the second truck loads two packages of $\{7, 3\}$ and the third truck loads a package of $\{9\}$, then the minimum value of the maximum load $P$ shall be 10.

Sample Input 2

4 2
1
2
2
6

Sample Output 2

6

If the first truck loads three packages of $\{1, 2, 2\}$ and the second truck loads a package of $\{6\}$, then the minimum value of the maximum load $P$ shall be 6.

Solution in C++14 (judge verdict: Time Limit Exceeded)

#include<bits/stdc++.h>
#define INF            1e9
#define REP(i,n)       for(lint i=0,i##_len=(n);i<i##_len;++i)
#define REP1(i,n)      for(lint i=1,i##_len=(n);i<=i##_len;++i)
#define REPI(i,ini,n)  for(lint i=(ini),i##_len=(n);i<i##_len;++i)
#define REPR(i,n)      for(lint i=(n)-1;i>=0;--i)
#define REPR1(i,n)     for(lint i=(n);i>0;--i)
#define REPRI(i,ini,n) for(lint i=(ini)-1,i##_len=(n);i>=i##_len;--i)
#define REPC(i,obj)    for(auto i:obj)
#define R_UP(a,b)      (((a)+(b)-1)/(b))
#define ALL(obj)       (obj).begin(),(obj).end()
using namespace std;
using lint = long long;
template<typename T = lint>inline T in() { T x; cin >> x; return x; }
int get_minimum_k(int p_candidate, const vector<int>&w)
{
	int result = 1;
	int temp = 0;
	REPC(i, w)
	{
		if (temp + i <= p_candidate)temp += i;
		else
		{
			result++;
			temp = i;
		}
	}
	return result;
}
int binary_search_p_candidate(int first_w, int last_w, int k, const vector<int>&w)
{
	if (first_w > last_w)return max(first_w, last_w);
	int p_candidate = first_w + (last_w - first_w) / 2;
	int min_k = get_minimum_k(p_candidate, w);
	return min_k == k ? p_candidate : min_k < k ? binary_search_p_candidate(first_w, p_candidate - 1, k, w) : binary_search_p_candidate(p_candidate + 1, last_w, k, w);
}
int lin_search_p(int p_candidate, int max_w, int k, const vector<int>&w)
{
	REPRI(i, p_candidate, max_w)if (get_minimum_k(i, w) > k)return i + 1;
	return max_w;
}
signed main()
{
	int n = in(), k = in();
	vector<int>w(n);
	int min_w = 10001, max_w = -1, sum_w = 0;
	REPC(&i, w)
	{
		i = in();
		if (min_w > i)min_w = i;
		if (max_w < i)max_w = i;
		sum_w += i;
	}
	int p_candidate = binary_search_p_candidate(min_w, sum_w, k, w);
	cout << lin_search_p(p_candidate, max_w, k, w) << endl;
}