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 Rust (judge verdict: Accepted)

use std::io;
use std::io::prelude::*;
use std::u32;

/// トラックk台のそれぞれの最大積載量がpの時、それが実現可能か判定する
fn check(vec: &Vec<u64>, k: u32, p: usize) -> bool {
    // k台のトラックでそれぞれ、荷物をpを超えるまで合計していく。
    // すべてのトラックについて合計が終わったとき、荷物が余っていればfalse, 荷物を載せきれていればtrueを返す

    let n = vec.len();
    let mut i = 0;
    for _ in 0..k {
        // トラック1台ごとに,iから荷物を載せていき、pを超えないところまで載せる
        let mut cur_w = 0;
        while i < n && (cur_w + vec[i]) as usize <= p {
            cur_w += vec[i];
            i += 1;
        }
    }

    // 添え字がnに達するなら末尾まで到達したので載せきることが出来た
    i >= n
}

/// k台のトラックにvecの荷物(n個)を割り当てる。この時トラックの最大積載量Pが最小になる値を求める
/// なお、割り当てにおいては、12|3456|のように、連続する荷物でないと同時に載せられない。
/// 場合の数だと、n-1 C k 通りの組合せが存在するはず。
fn calc_p(k: u32, vec: &Vec<u64>) -> u32 {
    // 二分法を用いるのがいいらしい

    let mut st: usize = 0;
    let mut end: usize = 100000 * 10000 + 1; // 荷物の最大個数*最大重量でpの取りうる最大値
    let mut p: usize = 0;
    while st < end {
        let mid = (st + end) / 2;   // 判定対象のpの計算
        if check(vec, k, mid) {
            p = mid;
            end = mid;
        } else {
            st = mid + 1;
        }
    }

    p as u32
}

fn main() {
    let sin = io::stdin();
    let mut l_iter = sin.lock().lines();

    let line = l_iter.next().unwrap().unwrap();
    let mut ws = line.split_whitespace();
    let n: usize = ws.next().unwrap().parse().unwrap();
    let k: u32 = ws.next().unwrap().parse().unwrap();

    let mut vec: Vec<u64> = vec![0; n];
    for i in 0..n {
        let line = l_iter.next().unwrap().unwrap();
        vec[i] = line.trim().parse().unwrap();
    }

    let res = calc_p(k, &vec);

    println!("{}", res);
}