#include <bits/stdc++.h>
using namespace std;

//高速化 
struct ponjuice{ponjuice(){cin.tie(0);ios::sync_with_stdio(0);cout<<fixed<<setprecision(20);}}PonJuice;
#define endl '\n' //インタラクティブ問題の時は消す

//型
using ll = long long;
using ld = long double;

// for文
#define overload4(a, b, c, d, e, ...) e
#define rep1(n)             for(ll i = 0; i < n; i++)
#define rep2(i, n)          for(ll i = 0; i < n; i++)
#define rep3(i, a, b)       for(ll i = a; i < b; i++)
#define rep4(i, a, b, step) for(ll i = a; i < b; i+= step)
#define rep(...) overload4(__VA_ARGS__, rep4, rep3, rep2, rep1)(__VA_ARGS__)
#define per1(n)             for(ll i = n-1; i >= 0; i--)
#define per2(i, n)          for(ll i = n-1; i >= 0; i--)
#define per3(i, a, b)       for(ll i = b-1; i >= a; i--)
#define per4(i, a, b, step) for(ll i = b-1; i >= a; i-= step)
#define per(...) overload4(__VA_ARGS__, per4, per3, per2, per1)(__VA_ARGS__)

//関数
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
template<class S, class T>inline bool chmax(S& a, T b){return a < b && ( a = b , true);}
template<class S, class T>inline bool chmin(S& a, T b){return a > b && ( a = b , true);}

//定数
constexpr ll mod = 998244353;
constexpr ll minf=-(1<<29);
constexpr ll inf=(1<<29);
constexpr ll MINF=-(1LL<<60);
// constexpr ll INF=(1LL<<60);
const int dx[4] ={-1, 0, 1, 0};
const int dy[4] ={ 0, 1, 0,-1};
const int dx8[8] ={-1,-1,-1, 0, 1, 1, 1, 0};
const int dy8[8] ={-1, 0, 1, 1, 1, 0,-1,-1};

void solve();
int main() {
	int t = 1;
    // cin >> t;
    while(t--)solve();
}

#include<atcoder/segtree>
using namespace atcoder;

using S = __int128_t;
S INF = (S)1'000'000'000'000'000'000'000'000'000'000'000'000;
S op(S a, S b) {
    if(INF / a <= b) return INF;
    return a * b;
}
S e() {return 1;}

void solve(){
    int n;
    cin >> n;
    n++;
    vector<ll> a(n);
    rep(i,0,n) cin >> a[i];
    segtree<S,op,e> seg(n);
    rep(i,0,n) {
        seg.set(i, a[i]);
    }

    vector<ll> dp(n+2);
    dp[0] = 1;
    dp[1] = -1;
    rep(i,0,n) {
        ll r = seg.max_right(i+1, [&](S x) {
            return a[i] % x == 0;
        });
        dp[r+1] = (dp[r+1] - dp[i]) % mod;
        dp[i+1] = (dp[i+1] + dp[i]*2) % mod;
    }

    cout << (dp[n]%mod+mod)%mod << endl;
}