1 //minamoto
 2 #include<iostream>
 3 #include<cstdio>
 4 #include<cmath>
 5 #define ll long long
 6 #define inf 0x3f3f3f3f3f3f3f3f
 7 using namespace std;
 8 #define getc() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++)
 9 char buf[1<<21],*p1=buf,*p2=buf;
10 template<class T>inline bool cmin(T&a,const T&b){return a>b?a=b,1:0;}
11 inline int read(){
12     #define num ch-'0'
13     char ch;bool flag=0;int res;
14     while(!isdigit(ch=getc()))
15     (ch=='-')&&(flag=true);
16     for(res=num;isdigit(ch=getc());res=res*10+num);
17     (flag)&&(res=-res);
18     #undef num
19     return res;
20 }
21 const int N=400005;const double Pi=acos(-1.0);
22 struct complex{
23     double x,y;
24     complex(double xx=0,double yy=0){x=xx,y=yy;}
25     inline complex operator +(complex b){return complex(x+b.x,y+b.y);}
26     inline complex operator -(complex b){return complex(x-b.x,y-b.y);}
27     inline complex operator *(complex b){return complex(x*b.x-y*b.y,x*b.y+y*b.x);}
28 }A[N],B[N];
29 int n,m,l,r[N],limit=1,a[N],b[N];ll a1,a2,b1,b2,ans=inf;
30 void FFT(complex *A,int type){
31     for(int i=0;i<limit;++i)
32     if(i<r[i]) swap(A[i],A[r[i]]);
33     for(int mid=1;mid<limit;mid<<=1){
34         complex Wn(cos(Pi/mid),type*sin(Pi/mid));
35         for(int R=mid<<1,j=0;j<limit;j+=R){
36             complex w(1,0);
37             for(int k=0;k<mid;++k,w=w*Wn){
38                 complex x=A[j+k],y=w*A[j+mid+k];
39                 A[j+k]=x+y,A[j+mid+k]=x-y;
40             }
41         }
42     }
43 }
44 int main(){
45 //    freopen("testdata.in","r",stdin);
46     n=read(),m=read();
47     for(int i=1;i<=n;++i)
48     a[i]=read(),a1+=a[i]*a[i],a2+=a[i];
49     for(int i=1;i<=n;++i)
50     b[i]=read(),b1+=b[i]*b[i],b2+=b[i];
51     for(int i=1;i<=n;++i)
52     A[i].x=A[i+n].x=a[i],B[i].x=b[n-i+1];
53     while(limit<=(n*3)) limit<<=1,++l;
54     for(int i=0;i<limit;++i)
55     r[i]=(r[i>>1]>>1)|((i&1)<<(l-1));
56     FFT(A,1),FFT(B,1);
57     for(int i=0;i<limit;++i) A[i]=A[i]*B[i];
58     FFT(A,-1);
59     for(int i=0;i<limit;++i) A[i].x=(ll)(A[i].x/limit+0.5);
60     for(int i=1;i<=n;++i)
61     for(int j=-m;j<=m;++j)
62     cmin(ans,a1+b1+j*j*n+2ll*j*(a2-b2)-2ll*(ll)A[n+i].x);
63     printf("%lld\n",ans);
64     return 0;
65 }