import sys
input = lambda: sys.stdin.readline().rstrip()
ii = lambda: int(input())
mi = lambda: map(int, input().split())
li = lambda: list(mi())
INF = 2**63-1
mod = 998244353
class fenwick_tree():
    n=1
    data=[0 for i in range(n)]
    def __init__(self,N):
        self.n=N
        self.data=[0 for i in range(N)]

    def add(self,p,x):
        assert 0<=p<self.n,"0<=p<n,p={0},n={1}".format(p,self.n)

        p+=1
        while(p<=self.n):
            self.data[p-1]+=x
            p += p&(-p)
    def sum(self,l,r):
        assert (0<=l and l<=r and r<=self.n),"0<=l<=r<=n,l={0},r={1},n={2}".format(l,r,self.n)
        return self.sum0(r)-self.sum0(l)
    def sum0(self,r):
        s=0
        while(r>0):
            s+=self.data[r-1]
            r-=r&-r
        return s

    def __getitem__(self, p):
        if isinstance(p, int):
            return self.sum(p, p + 1)
        else:
            return self.sum(p.start, p.stop)

    def __setitem__(self, p, x):
        return self.add(p, x - self[p])

    

n = ii()

p = li()

#転倒数の期待値を求めればよい．確率1/2で各間に仕切りが入れられるとして，
#P_i > P_j なる転倒が残る条件はP_iとP_jの間に仕切りがあること．
#よって，これの寄与は 1 - 1/2^(j - i) = 1 - 1/2^j * 2^i となる．
#bitを持ってえいえいやればよい

pow2 = [1] * (n + 1)

for i in range(1, n + 1):
    pow2[i] = (pow2[i - 1] * 2) % mod

inv2 = [1] * (n + 1)
i2 = pow(2, mod - 2, mod)
for i in range(1, n + 1):
    inv2[i] = (inv2[i - 1] * i2) % mod
BIT1 = fenwick_tree(n + 1)
BIT2 = fenwick_tree(n + 1)
ans = 0
for i in range(n):
    ans += BIT1[p[i]+1:n+1] - inv2[i] * BIT2[p[i]+1:n+1]
    ans %= mod
    BIT1[p[i]] += 1
    BIT2[p[i]] += pow2[i]

print(ans * pow2[n - 1] % mod)