Problem:
Score : 300 points

Problem Statement

We will play a one-player game using a number line and N pieces.

First, we place each of these pieces at some integer coordinate.

Here, multiple pieces can be placed at the same coordinate.

Our objective is to visit all of the M coordinates X_1, X_2, ..., X_M with these pieces, by repeating the following move:

Move: Choose a piece and let x be its coordinate. Put that piece at coordinate x+1 or x-1.

Note that the coordinates where we initially place the pieces are already regarded as visited.

Find the minimum number of moves required to achieve the objective.

Constraints

All values in input are integers.

1 \leq N \leq 10^5

1 \leq M \leq 10^5

-10^5 \leq X_i \leq 10^5

X_1, X_2, ..., X_M are all different.

Input

Input is given from Standard Input in the following format:

N M
X_1 X_2 ... X_M

Output

Find the minimum number of moves required to achieve the objective.

Sample Input 1

2 5
10 12 1 2 14

Sample Output 1

5

The objective can be achieved in five moves as follows, and this is the minimum number of moves required.

Initially, put the two pieces at coordinates 1 and 10.

Move the piece at coordinate 1 to 2.

Move the piece at coordinate 10 to 11.

Move the piece at coordinate 11 to 12.

Move the piece at coordinate 12 to 13.

Move the piece at coordinate 13 to 14.

Sample Input 2

3 7
-10 -3 0 9 -100 2 17

Sample Output 2

19

Sample Input 3

100 1
-100000

Sample Output 3

0

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

using namespace std;
#include<bits/stdc++.h>
#define BEGIN ios_base::sync_with_stdio(0);cin.tie(0);
#define END return EXIT_SUCCESS;
#define FOR(I,A,B) for((I)=(A);(I)<(B);(I)++)
#define REP(I,N) FOR((I),0,(N))
#define UP(I,A,B) for((I)=(A);(I)<=(B);(I)++)
#define DW(I,A,B) for((I)=(A);(I)>=(B);(I)--)
#define IN(P) cin>>(P)
#define IN2(P1,P2) cin>>(P1)>>(P2)
#define IN3(P1,P2,P3) cin>>(P1)>>(P2)>>(P3)
#define INS(I,N,V) REP((I),(N)) cin>>V[I]
#define INS2(I,N,V1,V2) REP((I),(N)) cin>>V1[I]>>V2[I];
#define INS3(I,N,V1,V2,V3) REP((I),(N)) cin>>V1[I]>>V2[I]>>V3[I];
#define OUT(P) cout<<(P)<<endl
#define OUT2(P1,P2) cout<<(P1)<<" "<<(P2)<<endl
#define OUT3(P1,P2,P3) cout<<(P1)<<" "<<(P2)<<" "<<(P3)<<endl
#define OUTS(I,N,V) REP((I),(N)) cout<<(V[i])<<endl
#define ALL(C) (C).begin(),(C).end()
#define RALL(C) (C).rbegin(),(C).rend()
#define mp make_pair
#define pb push_back
#define lb lower_bound
#define ub upper_bound
template<class T>inline bool chmax(T& a,T b){if(a<b){a=b;return true;}return false;}
template<class T>inline bool chmin(T& a,T b){if(a>b){a=b;return true;}return false;}
template<class T>inline T gcd(T a,T b){return b?gcd(b,a%b):a;}
template<class T>inline T lcm(T a,T b){return a*b/gcd(b,a%b);}
long N,M,i,ans;
vector<long> X,D;
inline void solve(){
  IN2(N,M);
  X.resize(M);
  INS(i,M,X);
  if(N>=M){
    OUT(0);
    return;
  }
  sort(ALL(X));
  D.resize(M-1);
  REP(i,M-1){
    D[i]=X[i+1]-X[i];
  }
  sort(ALL(D));
  REP(i,M-N){
    ans+=D[i];
  }
  OUT(ans);
}

int main(int argc,char**argv){
  BEGIN
  solve();
  END
}