传送门：
题解：
把a排序后，直接统计答案恰好为a[i]并不好做，可以统计答案>a[i]的方案数，设为\(f[i]\)。
即不存在一个联通块，所有的权值都<=a[i]。
那么如果枚举j个在i之前的点，分成k个联通块，容斥系数是\((-1)^k\)，选择系数\(C_i^j\)，剩下的边随便乱选，\(2^{(n-j)*(n-j-1)/2}\)。
设\(g[j]\)表示j个点，若有k个联通块，系数\((-1)^k\)，的所有方案系数和。
\(f[i]=\sum_{j=0}^iC_{i}^j*2^{(n-j)*(n-j-1)/2}*g[j]\)
求出\(g\)之后，卷一下就可以得到f，设G为g的EGF。
设\(a[i]\)表示i个点的简单无向图的方案数，A是它的EGF。
设\(b[i]\)表示i个点的简单无向连通图的方案数，B是它的EGF。
显然有\(e^B=A\)，又有\(e^B=G\)，所以\(G=A^{-1}\)，妙不可言。
Code：
#include<bits/stdc++.h>
#define fo(i, x, y) for(int i = x, B = y; i <= B; i ++)
#define ff(i, x, y) for(int i = x, B = y; i < B; i ++)
#define fd(i, x, y) for(int i = x, B = y; i >= B; i --)
#define ll long long
#define pp printf
#define hh pp("\n")
using namespace std;
const int mo = 998244353;
ll ksm(ll x, ll y) {
ll s = 1;
for(; y; y /= 2, x = x * x % mo)
if(y & 1) s = s * x % mo;
return s;
}
#define V vector<ll>
#define si size()
#define pb push_back
#define re resize
namespace ntt {
const int nm = 1 << 20;
int r[nm]; ll a[nm], b[nm], w[nm];
void bd() {
for(int n = 1; n < nm; n *= 2) {
w[n] = 1; ll v = ksm(3, (mo - 1) / 2 / n);
ff(j, 1, n) w[n + j] = w[n + j - 1] * v % mo;
}
}
void dft(ll *a, int n, int f) {
ff(i, 0, n) {
r[i] = r[i / 2] / 2 + (i & 1) * (n / 2);
if(i < r[i]) swap(a[i], a[r[i]]);
} ll b;
for(int i = 1; i < n; i *= 2) for(int j = 0; j < n; j += 2 * i) ff(k, 0, i)
b = a[i + j + k] * w[i + k], a[i + j + k] = (a[j + k] - b) % mo, a[j + k] = (a[j + k] + b) % mo;
if(f == -1) {
reverse(a + 1, a + n);
b = ksm(n, mo - 2);
ff(i, 0, n) a[i] = (a[i] + mo) * b % mo;
}
}
void dft(V &p, int f) {
int n = p.si;
ff(i, 0, n) a[i] = p[i];
dft(a, n, f);
ff(i, 0, n) p[i] = a[i];
}
V operator * (V p, V q) {
int n0 = p.si + q.si - 1, n = 1;
while(n < n0) n *= 2;
ff(i, 0, n) a[i] = b[i] = 0;
ff(i, 0, p.si) a[i] = p[i];
ff(i, 0, q.si) b[i] = q[i];
dft(a, n, 1); dft(b, n, 1);
ff(i, 0, n) a[i] = a[i] * b[i] % mo;
dft(a, n, -1);
p.re(n0);
ff(i, 0, n0) p[i] = a[i];
return p;
}
}
using ntt :: operator *;
using ntt :: dft;
V qni(V a) {
V b; b.re(1); b[0] = ksm(a[0], mo - 2);
for(int n = 1; n < a.si * 2; n *= 2) {
V c = a; c.re(n); c.re(2 * n);
V d = b; d.re(2 * n);
dft(d, 1); dft(c, 1);
ff(i, 0, 2 * n) d[i] = d[i] * d[i] % mo * c[i] % mo;
dft(d, -1);
b.re(n);
ff(i, 0, n) b[i] = (b[i] * 2 - d[i] + mo) % mo;
}
return b;
}
const int N = 5e5 + 5;
ll fac[N], nf[N];
int n, a[N];
V p, q;
int main() {
ntt :: bd();
scanf("%d", &n);
fo(i, 1, n) scanf("%d", &a[i]);
fac[0] = 1; fo(i, 1, n) fac[i] = fac[i - 1] * i % mo;
nf[n] = ksm(fac[n], mo - 2); fd(i, n, 1) nf[i - 1] = nf[i] * i % mo;
sort(a + 1, a + n + 1);
p.re(n + 1); fo(i, 0, n) p[i] = ksm(2, (ll) i * (i - 1) / 2) * nf[i] % mo;
p = qni(p);
fo(i, 0, n) p[i] = p[i] * ksm(2, (ll) (n - i) * (n - i - 1) / 2) % mo;
q.re(n + 1);
fo(i, 0, n) q[i] = nf[i];
p = p * q;
fo(i, 0, n) p[i] = p[i] * fac[i] % mo;
ll ans = 0;
fo(i, 1, n) ans = (ans + (ll) a[i] * (p[i - 1] - p[i] + mo)) % mo;
pp("%lld\n", ans);
}