#include <iostream>
#include <vector>
#include <cmath>
#include <map>
#include <set>
#include <iomanip>
#include <queue>
#include <algorithm>
#include <numeric>
#include <deque>
#include <complex>
#include <cassert>

using namespace std;
using ll = long long;

const ll modc = 998244353;
class mint {
    ll x;
public:
    mint(ll x=0) : x((x%modc+modc)%modc) {}
    mint operator-() const {
      return mint(-x);
    }
    mint& operator+=(const mint& a) {
        if ((x += a.x) >= modc) x -= modc;
        return *this;
    }
    mint& operator-=(const mint& a) {
        if ((x += modc-a.x) >= modc) x -= modc;
        return *this;
    }
    mint& operator*=(const  mint& a) {
        (x *= a.x) %= modc;
        return *this;
    }
    mint operator+(const mint& a) const {
        mint res(*this);
        return res+=a;
    }
    mint operator-(const mint& a) const {
        mint res(*this);
        return res-=a;
    }
    mint operator*(const mint& a) const {
        mint res(*this);
        return res*=a;
    }
    mint pow(ll t) const {
        if (!t) return 1;
        mint a = pow(t>>1);
        a *= a;
        if (t&1) a *= *this;
        return a;
    }
    mint inv() const {
        return pow(modc-2);
    }
    mint& operator/=(const mint& a) {
        return (*this) *= a.inv();
    }
    mint operator/(const mint& a) const {
        mint res(*this);
        return res/=a;
    }
    bool operator == (const mint& a) const{
        return x == a.x;
    }
    friend ostream& operator<<(ostream& os, const mint& m){
        os << m.x;
        return os;
    }
    friend istream& operator>>(istream& ip, mint&m) {
        ip >> m.x;
        return ip;
    }
};

int main(){

    ll N;
    mint ans=1;
    cin >> N;
    vector<pair<ll, ll>> P(N);
    vector<ll> L(N), R(N);
    for (int i=0; i<N; i++){
        cin >> P[i].first;
        P[i].second = i;
    }
    sort(P.begin(), P.end());
    L[0] = P[0].second; R[0] = P[0].second;
    for (int i=1; i<N; i++) L[i] = min(P[i].second, L[i-1]);
    for (int i=1; i<N; i++) R[i] = max(P[i].second, R[i-1]);
    for (int i=1; i<N; i++){
        if (L[i] == L[i-1] && R[i] == R[i-1]) ans *= (R[i]-L[i]+1-i);
    }
    cout << ans << endl;
}