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#include <bits/stdc++.h>
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using namespace std;
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const int N = 1000010;
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#define int long long
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#define endl "\n"
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int a[N], s[N];
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int n, k; // n个城市，传送半径为k
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int res;
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signed main() {
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cin >> n >> k; // n个城市，传送半径为k
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for (int i = 1; i < n; i++) { // 以起点为线段命名的编号,所以1~n共n个城市,其实是n-1条线段
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cin >> a[i]; // 走这条路要耗费时间a[i]
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s[i] += s[i - 1] + a[i]; // 前缀和
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}
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/*
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1、找出[1,n-1]的原始数组中，长度为k的子区间，加法和最大的区间段
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2、注意：如果是枚举原始数组,那么就是 [1,n-k]是有效区间
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3、现在遍历的是前缀和数组，为了表示a[1],需要使用的是s[1]-s[0],所以，i从0开始
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*/
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for (int i = 0; i + k < n; i++) res = max(res, s[i + k] - s[i]);
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cout << s[n - 1] - res << endl;
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}
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