题目链接：https://cn.vjudge.net/problem/183459/origin

题目

Professor GukiZ is concerned about making his way to school, because massive piles of boxes are blocking his way.

In total there are n piles of boxes, arranged in a line, from left to right, i-th pile (1 ≤ i ≤ n) containing ai boxes. Luckily, m students are willing to help GukiZ by removing all the boxes from his way. Students are working simultaneously. At time 0, all students are located left of the first pile. It takes one second for every student to move from this position to the first pile, and after that, every student must start performing sequence of two possible operations, each taking one second to complete. Possible operations are:

1. If i ≠ n, move from pile i to pile i + 1;
2. If pile located at the position of student is not empty, remove one box from it.

GukiZ's students aren't smart at all, so they need you to tell them how to remove boxes before professor comes (he is very impatient man, and doesn't want to wait). They ask you to calculate minumum time t in seconds for which they can remove all the boxes from GukiZ's way. Note that students can be positioned in any manner after t seconds, but all the boxes must be removed.

Input

The first line contains two integers n and m (1 ≤ n, m ≤ 105), the number of piles of boxes and the number of GukiZ's students.

The second line contains n integers a1, a2, ... an (0 ≤ ai ≤ 109) where ai represents the number of boxes on i-th pile. It's guaranteed that at least one pile of is non-empty.

Output

In a single line, print one number, minimum time needed to remove all the boxes in seconds.

Examples

Input

2 1
1 1

Output

4

Input

3 2
1 0 2

Output

5

Input

4 100
3 4 5 4

Output

5

Note

First sample: Student will first move to the first pile (1 second), then remove box from first pile (1 second), then move to the second pile (1 second) and finally remove the box from second pile (1 second).

Second sample: One of optimal solutions is to send one student to remove a box from the first pile and a box from the third pile, and send another student to remove a box from the third pile. Overall, 5 seconds.

Third sample: With a lot of available students, send three of them to remove boxes from the first pile, four of them to remove boxes from the second pile, five of them to remove boxes from the third pile, and four of them to remove boxes from the fourth pile. Process will be over in 5 seconds, when removing the boxes from the last pile is finished.

题意：给你两个数n,m;这个路的长路，和学生人数，第二行给出每个单位长度的箱子数，每个学生可以在一秒内向前走一步或者搬掉一个箱子。问把箱子搬完最少要多少时间。

思路：用二分法，把所需要的时间二分，二分时间为x,那么每个学生就都有x秒时间，可以让学生一个一个去般，如果一个学生可以在x秒内搬完所有的箱子，那么时间足够用，如果一个学生不能搬完所有箱子，那么会再派一个人去到前一个没有办完的箱子地方，接着搬，直到人数用完，那么时间不够用。

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<algorithm>
using namespace std;
long long  a[100005];
int n,m;
 long long  t=0;
    int w=0;
int fin(long long  x)
{
long long e=0;
     int ren=m;
    for(int i=1;i<=w;i++)
    {

        e+=a[i];
        while(e+i>=x)
        {
            e-=x-i;
            ren--;
            if(ren<0)
            {
                return 0;
            }
        }
    }
    if(ren==0)
    {
      if(e>0)
      {
          return 0;
      }

    }
    return 1;
}
int main()
{
    long long sum=0,l,r;
    scanf("%d %d",&n,&m);

    for(int i=1; i<=n; i++)
    {
        scanf("%lld",&a[i]);
        sum+=a[i];
        if(a[i])
        {
            w=i;
        }
    }
    l=w;r=w+sum;
    while(l<=r)
    {
        long long mid=(l+r)/2;
        if(fin(mid)==1)
        {
           t=mid;
           r=mid-1;
        }
        else
        {
            l=mid+1;
        }
    }

    printf("%lld\n",t);
    return 0;
}