#include <bits/stdc++.h>
// #include <atcoder/all>
using namespace std;
// using namespace atcoder;
#define rep2(i,m,n) for (int i = (m); i < (n); ++i)
#define rep(i,n) rep2(i,0,n)
typedef long long int ll;
typedef long double ld;
typedef pair<int, int> P;
template<typename T> struct V : vector<T> { using vector<T>::vector; };
V() -> V<int>;
V(size_t) -> V<int>;
template<typename T> V(size_t, T) -> V<T>;
template<typename T> vector<T> make_vec(size_t n, T a) { return vector<T>(n, a); }
template<typename... Ts> auto make_vec(size_t n, Ts... ts) { return vector<decltype(make_vec(ts...))>(n, make_vec(ts...)); }
template<typename T> ostream &operator<<(ostream &os, const vector<T> &v) { for (auto &e : v) os << e << ' '; return os; }
struct fast_ios { fast_ios(){ cin.tie(nullptr); ios::sync_with_stdio(false); cout << fixed << setprecision(20); }; } fast_ios_;
template<typename T> inline bool chmin(T &a, T b) { if (a > b) { a = b; return true; } return false; }
template<typename T> inline bool chmax(T &a, T b) { if (a < b) { a = b; return true; } return false; }
const int INF = 1<<30;
const ll LINF = 1LL<<61;
// const ll MOD = 1000000007;
const ll MOD = 998244353;
// using mint = modint1000000007;
int  MAX = 3000;

int main()
{
    int n;
    cin >> n;
    vector<ll> a(n);
    rep(i,n) cin >> a[i];
    sort(a.begin(), a.end());
    vector<ll> as(n);
    for(int i=n-1; i>1; i--) as[i-2] = abs(a[i] - a[i-1]);
    for(int i=n-1; i>0; i--) chmax(as[i-1], as[i]);
    cout << as << endl;
    // i番目までみた, 最後に使ったのがj, スコアの総和
    auto dp = make_vec(n+1, n+1, 0LL);
    auto done = make_vec(n+1, MAX, false);

    auto rec = [&](auto self, int x, int y) -> ll
    {
        if(done[x][y]) return dp[x][y];
        if(x <= 0) return dp[x][y];
        for(int j = y-1; j >= 0; j--)
        {
            if(x > 0)dp[x][y] = (dp[x][y] + self(self, x-1, j) + max(abs(a[x-1]-a[j]), as[x-1])) % MOD;
        }
        done[x][y] = true;
        return dp[x][y] % MOD;
    };
    ll ans = 0;
    rep(i,n+1) ans = (ans + rec(rec, n, i) % MOD);
    cout << ans << endl;
    // rep(i,n+1) cout << dp[i] << endl;

    return 0;
}