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++ (GCC 9.2.1) (judge verdict: Accepted)

//s配列に格納した単語を、辞書順にソートする
//  sort(s.begin(), s.end());
//  string t = "keyence";//で文字列入力ができる
//s.empty() //emptyなら1を出力 入っていれば0を出力
/*//ABC018-B 部分的にreverseをかける解法
int main() {
  string s;
  cin >> s;
  int n;
  cin >> n;
  vector<int>a(n);
  vector<int>b(n);
  rep(i,n) cin>>a.at(i)>>b.at(i);
  rep(i,n)a.at(i)--;
  rep(i,n)b.at(i)--;
  string t;
  rep(i,n){
    t = s;
    for(int k=0;k<=b.at(i)-a.at(i);k++){
      t.at(a.at(i)+k) = s.at(b.at(i)-k);
    }
    s = t;
  }
  cout << s << endl;
}
*///ABC018-B
//  string s;
//  cin >> s;
//  cout << s.at(0) <<endl;
//  cout << s.at(0)-'0' <<endl;
//  Aが16？
//  cout << char(i+48) << endl;//なぜかaは47と得る時がある。+48で出力もaにできる。
//  cout << char(97) << endl;//アスキーコードでaを出力
// sort(b.begin(), b.end());//bという配列を小さい方からソート
// reverse(b.begin(), b.end());//bという配列をリターン
// 10桁でdouble出力するパターン１ 
// printf("%.10f\n", ma);
// 10桁でdouble出力するパターン２
//cout << fixed << setprecision(10);
//  cout << ma << endl;

/*01 02 03 12 13 23　と６回見ていくパターン 
for(int i=0;i<n-1;i++){
  for(int j=i+1;j<n;j++){
    }
  }
*///vector<vector<int>> data(3, vector<int>(4));//int型の2次元配列(3×4要素の)の宣言
//int64_t a;
//10のi乗pow(10, i);
/*string s; stringでの文字列を数字型に変える方法
  cin >> s;
  rep(i,s.size()-2) {
  int a= (s.at(i)-'0')*100 + (s.at(i+1)-'0')*10+ s.at(i+2) -'0';
    */
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
int kan_hyaku(int n){
  int kurai = 0;
  for(int i=0;i<3;i++){
    kurai = n%10;
    n /=10;
  }
  return kurai;
}
int kan_ju(int n){
  int kurai = 0;
  for(int i=0;i<2;i++){
    kurai = n%10;
    n /=10;
  }
  return kurai;
}
int kan_ichi(int n){
  int kurai = 0;
  for(int i=0;i<1;i++){
    kurai = n%10;
    n /=10;
  }
  return kurai;
}
int main() {
  ll n,m;
  cin >> n >> m;
  vector<ll>x(m);
  rep(i,m) cin >> x.at(i);
  if(n > m ){
    n = m;
  }
  if(m==1){
    cout << 0 << endl;
    return 0;
  }
  vector<ll>b(m-1);
  sort(x.begin(),x.end());
  for(int i=0;i<m-1;i++){
    b.at(i) = x.at(i+1) - x.at(i);
  }
  sort(b.begin(),b.end());
  reverse(b.begin(),b.end());
  for(int i=0;i<n-1;i++){
    b.at(i) = 0;
  }
  ll sum = 0;
  rep(i,m-1) sum += b.at(i);
  cout << sum << endl;
}