#include <bits/stdc++.h>
#define int long long
using namespace std;
const int MAXN = 300;
const int MAXM = 1'000'000;
const int INF = MAXN * MAXM + 5;
int n, m;
int x[MAXN + 5];
// 一共喝 num 滴水，喝完了 l~r 的水，停在了 左0/右1 的最小总消耗
int f[MAXN + 5][MAXN + 5][2];
signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin >> n >> m;
    for (int i = 1; i <= n; i++)
        cin >> x[i];

    sort(x + 1, x + n + 1);
    int ans = 0;
    // 碰了 num 滴水
    for (int num = 1; num <= n; num++)
    {
        // 计算 f[l][r][0/1]
        for (int len = 1; len <= num; len++)
        {
            for (int l = 1; l + len - 1 <= n; l++)
            {
                int r = l + len - 1;
                f[l][r][0] = f[l][r][1] = INF;
                if (len == 1)
                {
                    // 所有水珠都在消耗
                    f[l][r][0] = f[l][r][1] = abs(x[l]) * num;
                    continue;
                }
                int rem = num - len + 1; // 还有几个在消耗
                // l+1 -> l
                f[l][r][0] = min(f[l][r][0],
                                 f[l + 1][r][0] + (x[l + 1] - x[l]) * rem);
                // r -> l
                f[l][r][0] = min(f[l][r][0],
                                 f[l + 1][r][1] + (x[r] - x[l]) * rem);
                // l -> r
                f[l][r][1] = min(f[l][r][1],
                                 f[l][r - 1][0] + (x[r] - x[l]) * rem);
                // r+1 -> r
                f[l][r][1] = min(f[l][r][1],
                                 f[l][r - 1][1] + (x[r] - x[r - 1]) * rem);
            }
        }
        for (int l = 1; l + num - 1 <= n; l++)
        {
            ans = max(ans, num * m - f[l][l + num - 1][0]);
            ans = max(ans, num * m - f[l][l + num - 1][1]);
        }
    }
    cout << ans;
    return 0;
}