#include <bits/stdc++.h>
#include <atcoder/modint>

using namespace std;
using ll = long long;
using namespace atcoder;
using mint = modint998244353;

//calculate floor(sqrt(s))
ll isqrt(ll s){
    ll l=0, r=3e9, c;
    while(r-l>1){
        c = (l+r)/2;
        if (c*c <= s) l = c;
        else r = c;
    }
    return l;
}

int main(){
    cin.tie(nullptr);
    ios_base::sync_with_stdio(false);

    /*
       sum(i=1, N)(M-(M/k)k)=MN-sum(i=1,N)(M/k)k
       M/iがsqrt(M-1)以下のiはO(sqrt(M))個しかないので愚直にやる。
       1<=j<=sqrt(M)に対して、M/k=jとなるkをまとめて
       j*(kの総和)
       j<=M/k<j+1

       M/(j+1)<k<=M/j
       L = floor(M/(j+1))+1
       R = max(floor(M/j), N)
    */
    ll N, M, K, L, R;
    cin >> N >> M;
    mint ans=0, S, iv=mint(2).inv();
    K = isqrt(M);
    for (int i=1; i<=1e9; i++){
        if (M/i <= K) break;
        ans += (M/i)*i;
    }
    for (int j=K; j>=1; j--){
        L = M/(j+1)+1;
        R = min(M/j, N);
        if (L>R) continue;
        //L~Rの和
        S = mint(R)*(R+1)*iv-mint(L)*(L-1)*iv;
        ans += S*j;
    }

    cout << (mint(N)*M-ans).val() << endl;

    return 0;
}