//O(NlogN)
#include <iostream>
#include <cstdio>
#include <vector>
#include <atcoder/fenwicktree>
#define int long long
using namespace std;
using namespace atcoder;

int mod = 998244353;
int n;
vector<int> p;
vector<int> fact, finv, inv;	//fact[n] = n!, finv[n] = n!の逆元, inv[n] = nの逆元

//参考: よくやる二項係数、https://drken1215.hatenablog.com/entry/2018/06/08/210000
void comb_init(int n) {
	fact.resize(n);
	finv.resize(n);
	inv.resize(n);
	fact[0] = fact[1] = 1;
	finv[0] = finv[1] = 1;
	inv[1] = 1;
	for (int i = 2; i < n; i++) {
		fact[i] = fact[i - 1] * i % mod;
		inv[i] = mod - inv[mod % i] * (mod / i) % mod;
		finv[i] = finv[i - 1] * inv[i] % mod;
	}
}

int comb(int n, int k) {
	if (n < 0 || k < 0 || n < k) return 0;
	return fact[n] * finv[k] % mod * finv[n - k] % mod;
}

signed main() {
	cin >> n;
	comb_init(n);
	
	p.resize(n);
	for (int i = 0; i < n; i++) { scanf("%lld", &p[i]); p[i]--; }
	
	fenwick_tree<int> fwL(n);
	vector<int> A(n), B(n);
	for (int i = 0; i < n; i++) {
		A[i] = fwL.sum(0, p[i]);
		B[i] = i - A[i];
		fwL.add(p[i], 1);
	}
	
	fenwick_tree<int> fwR(n);
	vector<int> C(n), D(n);
	for (int i = n - 1; i >= 0; i--) {
		C[i] = fwR.sum(0, p[i]);
		D[i] = n - 1 - i - C[i];
		fwR.add(p[i], 1);
	}
	
	int ans = 0;
	for (int i = 0; i < n; i++) {
		ans += comb(A[i] + D[i], A[i]) * comb(B[i] + C[i], B[i]) % mod;
		ans %= mod;
	}
	
	cout << ans << endl;
	return 0;
}