Coderfocers-551C

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 a_i boxes.
Luckily, m students are willing to help GukiZ by removing all the boxes from his way. Students are working simultaneously. At time0, 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 aftert 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, ... a_n (0 ≤ a_i ≤ 10⁹)
where a_i 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.

Example

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：在坐标上任意位置有任意的箱子，现在有M个人，每个人都有两个选择，如果该位置有箱子就搬箱子，要么就移动到下一个位置，这两个过程都花费1秒。求把所有箱子都移走所用的最下时间；

题解；用二分法对时间二分，判断是否可以移完所有箱子，直到找到恰好可以移走完所有箱子的时间，即为最短时间。

AC代码为：

#include<cstdio>

#include<iostream>

#include<cmath>

using namespace std;

const int mod = 1e9 + 7;

int n, m, tot;

long long  a[100005];

int f(long long x)

{
long long s=0,num=m;
for(int i=1;i<=tot;i++)
{
s+=a[i];
while(s+i>=x)
{
s-=x-i;
num--;
if(num<0)
return 0;
}
}
if (num == 0) return s <= 0;
return 1;

}

int main()

{
long long sum,left,mid,right,ans,flag;

while(~scanf("%d%d",&n,&m))
{
sum=0;
for(int i=1;i<=n;i++)
{
scanf("%lld",a+i);
sum+=a[i];
if(a[i])
tot=i;
}
ans=sum+tot;
left=tot,right=ans;

while(left<=right)
{
mid=(left+right)>>1;

if(f(mid))
{
flag=mid;
right=mid-1;
}
else
left=mid+1;

}
printf("%lld\n",flag);

}

return 0;

}