#include <iostream>
#include <atcoder/modint>
#include <atcoder/math>
#include <atcoder/segtree>

using namespace std;
using namespace atcoder;
using ll = long long;
using mint = modint998244353;
const int MX = 1000010;
mint f[MX],inv[MX],fi[MX];
constexpr ll mod = 998244353;
void solve(){
    inv[1] = 1;
    for(int i=2;i<MX;i++){
        inv[i] = mod - (mod/i)*inv[mod%i];
    }
    f[0] = fi[0] = 1;
    for(int i=1;i<MX;i++){
        f[i] = f[i-1]*i;
        fi[i] = fi[i-1]*inv[i];
    }
}

mint nck(ll n, ll k){
    if(n<0 || k<0 || n<k) return 0;
    return f[n]*fi[k]*fi[n-k];
}

mint pw(mint a,ll x){
    mint ret = 1;
    while(x){
        if(x&1) (ret *= a);
        (a *= a); x /= 2;
    }
    return ret;
}

mint op(mint a,mint b){return a + b;}
mint e(){return 0;}

const int B = 200;
ll N[50010],d[50010],cnt[50010];
ll di[50010];
int main(){
    solve();
    ll i,j,n,m; cin >> n >> m;
    for(i=0;i<n;i++) cin >> d[i];
    for(i=1;i<=m;i++) di[i] = m/i;
    for(i=0;i<n;i++) N[i] = di[d[i]];
    mint ans = 0;
    // (?,小)
    for(i=n - 1;i>=0;i--){
        for(j=1;j<=B;j++){
            ans += inv[N[i]]*inv[di[j]]*cnt[j]*floor_sum(N[i],j,d[i],d[i] - 1);
        }
        cnt[d[i]]++;
    }
    // debug
    mint ans1 = ans;
    for(i=0;i<n;i++) ans1 *= N[i];

    // (小,大)
    for(j=1;j<=m;j++) cnt[j] = 0;
    for(i=0;i<n;i++){
        if(d[i]>B){
            for(j=1;j<=B;j++){
                ans += inv[N[i]]*inv[di[j]]*cnt[j]*(m/j*N[i] - floor_sum(N[i],j,d[i],d[i]));
            }
        }
        cnt[d[i]]++;
    }
    // (大,大)
    segtree<mint,op,e> seg(m + 1);
    auto upd = [&](int pos,mint val){
        seg.set(pos,seg.get(pos) + val);
    };
    for(i=n - 1;i>=0;i--){
        if(d[i]<=B) continue;
        for(j=d[i];j<=m;j+=d[i]){
            ans += seg.prod(0,j)*inv[N[i]];
        }
        for(j=d[i];j<=m;j+=d[i]) upd(j,inv[N[i]]);
    }
    for(i=0;i<n;i++) ans *= N[i];
    cout << ans.val() << "\n";
}