1 #include <cmath>
  2 #include <cstdio>
  3 #include <cstring>
  4 #include <cstdlib>
  5 #include <iostream>
  6 #include <algorithm>
  7 
  8 #define ri register int
  9 
 10 #define lim 10000000
 11 
 12 char *c = new char[lim];
 13 
 14 template <class T>
 15 void read(T &x)
 16 {
 17     x = 0;
 18     
 19     while (*c < ‘0‘)++c;
 20     
 21     while (*c >= ‘0‘)
 22         x = x*10 + *c++ - ‘0‘;
 23 }
 24 
 25 template <class T>
 26 void Min(T &a, T b)
 27 {
 28     if (a > b)a = b;
 29 }
 30 
 31 template <class T>
 32 void Max(T &a, T b)
 33 {
 34     if (a < b)a = b;
 35 }
 36 
 37 #define N 1000005
 38 
 39 int n, m;
 40 int h[N];
 41 
 42 struct node
 43 {
 44     int lt, rt, cnt;
 45     long long sum;
 46 }tree[N * 6];
 47 
 48 void build(int p, int l, int r)
 49 {
 50     node &t = tree[p];
 51     
 52     t.lt = l;
 53     t.rt = r;
 54     
 55     t.cnt = t.sum = 0;
 56     
 57     if (l ^ r)
 58     {
 59         int mid = (l + r) >> 1;
 60         
 61         build(p << 1, l, mid);
 62         build(p << 1 | 1, mid + 1, r);
 63     }
 64 }
 65 
 66 void insert(int p, int pos, int val1, int val2)
 67 {
 68     node &t = tree[p];
 69     
 70     t.cnt += val1;
 71     t.sum += val2;
 72     
 73     if (t.lt ^ t.rt)
 74     {
 75         int mid = (t.lt + t.rt) >> 1;
 76         
 77         if (pos <= mid)
 78             insert(p << 1, pos, val1, val2);
 79         else
 80             insert(p << 1 | 1, pos, val1, val2);
 81     }
 82 }
 83 
 84 int query1(int p, int rank)
 85 {
 86     node &t = tree[p];
 87     
 88     if (t.lt == t.rt)return t.lt;
 89     
 90     if (tree[p << 1].cnt >= rank)
 91         return query1(p << 1, rank);
 92     else
 93         return query1(p << 1 | 1, rank - tree[p << 1].cnt);    
 94 }
 95 
 96 long long query2(int p, int l, int r)
 97 {
 98     if (l > r)return 0LL;
 99     
100     node &t = tree[p];
101     
102     if (l == t.lt && r == t.rt)
103         return t.sum;
104         
105     int mid = (t.lt + t.rt) >> 1;
106     
107     if (r <= mid)
108         return query2(p << 1, l, r);
109     if (l > mid)
110         return query2(p << 1 | 1, l, r);
111     return query2(p << 1, l, mid) + query2(p << 1 | 1, mid + 1, r);
112 }
113 
114 int query3(int p, int l, int r)
115 {
116     if (l > r)return 0LL;
117     
118     node &t = tree[p];
119     
120     if (l == t.lt && r == t.rt)
121         return t.cnt;
122         
123     int mid = (t.lt + t.rt) >> 1;
124     
125     if (r <= mid)
126         return query3(p << 1, l, r);
127     if (l > mid)
128         return query3(p << 1 | 1, l, r);
129     return query3(p << 1, l, mid) + query3(p << 1 | 1, mid + 1, r);
130 }
131 
132 signed main(void)
133 {
134     fread(c, 1, lim, stdin);
135     
136     read(n); 
137     read(m);
138     
139     ri maxi = 0;
140     ri mini = N;
141     
142     for (ri i = 1; i <= n; ++i)
143     {
144         read(h[i]);
145         
146         Min(mini, h[i]);
147         Max(maxi, h[i]);
148     }
149     
150     build(1, 0, N);
151     
152     for (ri i = 1; i < m; ++i)
153         insert(1, h[i], 1, h[i]);
154         
155     int d = (m + 1) >> 1;
156     
157     long long ans = 1e18 + 9;
158         
159     for (ri i = m; i <= n; ++i)
160     {
161         insert(1, h[i], 1, h[i]);
162         
163         {
164             int mid = d;
165             
166             long long q = query1(1, mid), res = 0LL;
167             
168             long long lc = query3(1, 0, q - 1);
169             long long rc = query3(1, q + 1, N);
170             
171             res += lc*q - query2(1, 0, q - 1);
172             res += query2(1, q + 1, N) - rc*q;
173             
174             Min(ans, res);
175         }
176         
177         insert(1, h[i - m + 1], -1, -h[i - m + 1]);
178     }
179     
180     printf("%lld\n", ans);
181 }