#include <iostream>
#include <queue>
using namespace std;
 
int main() {
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	int x, n;
	cin >> x >> n;
	priority_queue<int, vector<int>, greater<int>> PQ;
	for (int i = 0; i < n; i++) {
		int a;
		cin >> a;
		PQ.push(a);
	}
	long long ans = 0;
	for (int i = 1; i < n; i++) {
		int a = PQ.top();
		PQ.pop();
		int b = PQ.top();
		PQ.pop();
		PQ.push(a + b);
		ans += a + b;
	}
 
	cout << ans << "\n";
	return 0;
}

import java.io.*;
import java.util.*;
 
public class StickDivision {
	public static void main(String[] args) throws IOException {
		BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer(in.readLine());
 
		// with our algorithm the variable x will not be used at all
		int x = Integer.parseInt(st.nextToken());
		int n = Integer.parseInt(st.nextToken());
		Queue<Integer> sticks = new PriorityQueue<Integer>();
		st = new StringTokenizer(in.readLine());
		for (int i = 0; i < n; i++) sticks.add(Integer.valueOf(st.nextToken()));
 
		long costs = 0;
		// we keep combining two smallest sticks a and b
		// until there is only one stick remaining, which will have the length x
		while (sticks.size() > 1) {
			int a = sticks.remove();
			int b = sticks.remove();
			sticks.add(a + b);
			// the length of the combined stick is the cost for dividing them
			// into the separated sticks a and b
			costs += a + b;
		}
		System.out.println(costs);
	}
}

import heapq
 
x, n = map(int, input().split())
d = list(map(int, input().split()))
 
heapq.heapify(d)
res = 0
for _ in range(n - 1):
	a = heapq.heappop(d)
	b = heapq.heappop(d)
	res += a + b
	heapq.heappush(d, a + b)
 
print(res)