Problem:
Score : 500 points

Problem Statement

On an xy plane, in an area satisfying 0 ≤ x ≤ W, 0 ≤ y ≤ H, there is one house at each and every point where both x and y are integers.

There are unpaved roads between every pair of points for which either the x coordinates are equal and the difference between the y coordinates is 1, or the y coordinates are equal and the difference between the x coordinates is 1.

The cost of paving a road between houses on coordinates (i,j) and (i+1,j) is p_i for any value of j, while the cost of paving a road between houses on coordinates (i,j) and (i,j+1) is q_j for any value of i.

Mr. Takahashi wants to pave some of these roads and be able to travel between any two houses on paved roads only. Find the solution with the minimum total cost.

Constraints

1 ≦ W,H ≦ 10^5

1 ≦ p_i ≦ 10^8(0 ≦ i ≦ W-1)

1 ≦ q_j ≦ 10^8(0 ≦ j ≦ H-1)

p_i (0 ≦ i ≦ W−1) is an integer.

q_j (0 ≦ j ≦ H−1) is an integer.

Input

Inputs are provided from Standard Input in the following form.

W H
p_0
:
p_{W-1}
q_0
:
q_{H-1}

Output

Output an integer representing the minimum total cost.

Sample Input 1

2 2
3
5
2
7

Sample Output 1

29

It is enough to pave the following eight roads.

Road connecting houses at (0,0) and (0,1)

Road connecting houses at (0,1) and (1,1)

Road connecting houses at (0,2) and (1,2)

Road connecting houses at (1,0) and (1,1)

Road connecting houses at (1,0) and (2,0)

Road connecting houses at (1,1) and (1,2)

Road connecting houses at (1,2) and (2,2)

Road connecting houses at (2,0) and (2,1)

Sample Input 2

4 3
2
4
8
1
2
9
3

Sample Output 2

60

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
int lub(int l,int r,int n,int *d){
  int m=(l+r)/2;
  if(l==r)return m;
  return n<=d[m]?lub(l,m,n,d):lub(m+1,r,n,d);
}
int R=1,C=1,H[2000010],N[2000010];
//評価関数（いまはMAX）
int hyouka(int a,int b){
  if(C<b)return 1;
  if(C<a||b==0)return 0;
  return N[H[a]]<N[H[b]]?1:0;
}
//挿入関数
void hin(int a){
  int i=C++;
  for(N[H[0]=R]=a;hyouka(0,i/2);i/=2)H[i]=H[i/2];
  H[i]=R++;
}
//取り出す関数
int hout(){
  int rt=H[1],i,j=2,k=H[--C];
  for(i=1;hyouka(i,C);i=j)H[i]=H[j=i*2+1-hyouka(i*2,i*2+1)];
  H[j/2]=k;
  return rt;
}
long long int MIN(long long a,long long b){return a<b?a:b;}
/*long long f(long long a,long long b,long long min,long long *t){
  long long i,ans=min;
  for(i=0;i<b;i++)ans+=MIN(min+t[i],t[i]*(a+1));//,printf("%lld\n",ans);
  return ans;
  }//*/
long long dx[100010],dy[100010];
long long f(int a,int b,long long min,int *s,int *t,long long *d){
  long long i,j;
  //for(i=0;i<a;i++)printf("%d ",t[i]);printf("\n"); 
  for(i=0;i<b;i++){
    j=lub(0,a,s[i],t);//printf("%d %lld\n",s[i],j);
    min+=d[j]+s[i]*(a+1-j);
  }
  return min;
}
    
int main(){
  int i,h,w,a,x[100010],y[100010];
  long long sw,sh;
  scanf("%d %d",&w,&h);
  for(i=sw=0;i<w;i++){
    scanf("%d",&a);
    hin(a);
    sw+=a;
  }
  for(i=0;i<w;i++)x[i]=N[hout()];
  for(i=dx[0]=0;i<w;i++)dx[i+1]=dx[i]+x[i];
  R=C=1;
  for(i=sh=0;i<h;i++){
    scanf("%d",&a);
    hin(a);
    sh+=a;
  }
  for(i=0;i<w;i++)y[i]=N[hout()];
  for(i=dy[0]=0;i<w;i++)dy[i+1]=dy[i]+y[i];
  printf("%lld\n",sh<sw?f(h,w,sh,x,y,dy):f(w,h,sw,y,x,dx));
  return 0;
}