B. Fence

B. Fence

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

There is a fence in front of Polycarpus's home. The fence consists of n planks of the same width which go one after another from left to right. The height of thei-th plank is h_i meters, distinct planks can have distinct heights.

Fence for n = 7 andh = [1, 2, 6, 1, 1, 7, 1]

Polycarpus has bought a posh piano and is thinking about how to get it into the house. In order to carry out his plan, he needs to take exactlyk consecutive planks from the fence. Higher planks are harder to tear off the fence, so Polycarpus wants to find suchk consecutive planks that the sum of their heights is minimal possible.

Write the program that finds the indexes of k consecutive planks with minimal total height. Pay attention, the fence is not around Polycarpus's home, it is in front of home (in other words, the fence isn't cyclic).

Input

The first line of the input contains integers n andk (1 ≤ n ≤ 1.5·10⁵, 1 ≤ k ≤ n) — the number of planks in the fence and the width of the hole for the piano. The second line contains the sequence of integers h₁, h₂, ..., h_n (1 ≤ h_i ≤ 100), where h_i is the height of thei-th plank of the fence.

Output

Print such integer j that the sum of the heights of planksj, j + 1, ...,j + k - 1 is the minimum possible. If there are multiple suchj's, print any of them.

Sample test(s)

Input

7 3
1 2 6 1 1 7 1

Output

3

Note

In the sample, your task is to find three consecutive planks with the minimum sum of heights. In the given case three planks with indexes 3, 4 and 5 have the required attribute, their total height is 8.

题意：求连续k个元素之和最小时开始的位置。

用例来解释

ACcode：

#include <iostream>
using namespace std;
long a[1500000],n,m;
int main()
{
    long i,j,min;
    cin>>n>>m;
    
	for(i=1;i<=n;i++)
	{
		cin>>a[i];
		a[i]+=a[i-1]; //将前面的每一项加到下一项
	}   
	min=0xfffffff;   //min为无穷大
	for(i=m;i<=n;i++)   //从第m个开始每次移一个，后面的m位
	{
		if(a[i]-a[i-m]<min)   //a[i]-a[i-m]为后3个值
		{
			min=a[i]-a[i-m];
			j=i-m+1;
		}
		
	}
	cout<<j<<endl;
	
    return 0;
}