#include <bits/stdc++.h>
#include <atcoder/all>
using namespace std;
using namespace atcoder;
//using mint = modint1000000007;
//const int mod = 1000000007;
using mint = modint998244353;
const int mod = 998244353;
//const int INF = 1e9;
//const long long LINF = 1e18;
#define rep(i, n) for (int i = 0; i < (n); ++i)
#define rep2(i,l,r)for(int i=(l);i<(r);++i)
#define rrep(i, n) for (int i = (n-1); i >= 0; --i)
#define rrep2(i,l,r)for(int i=(r-1);i>=(l);--i)
#define all(x) (x).begin(),(x).end()
#define allR(x) (x).rbegin(),(x).rend()
#define endl "\n"
#define P pair<int,int>
template<typename A, typename B> inline bool chmax(A & a, const B & b) { if (a < b) { a = b; return true; } return false; }
template<typename A, typename B> inline bool chmin(A & a, const B & b) { if (a > b) { a = b; return true; } return false; }
mint inv2, inv6;
mint f(int n) {
	//[0,n)
	mint x = n - 1;
	mint res = x * (x + 1) * inv2;
	return res;
}
mint f2(int n) {
	//[0,n)
	mint x = n - 1;
	mint res = x * (x + 1)*(2 * x + 1) * inv6;
	return res;
}
int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	int n, m; cin >> n >> m;
	int mid = sqrt(n);
	inv2 = ((mint)2).inv();
	inv6 = ((mint)6).inv();
	mint ans = (mint(m))*n *(n + 1) * inv2;
	rep2(i, 1, m + 1) {
		int x = n / i;
		if (x < mid)break;
		ans -= (mint(n + 1))*x*i;
		ans += (mint(x))*(x + 1)*i * i * inv2;
	}
	rep2(i, 1, mid) {
		int l = 1 + n / (i + 1);
		int r = min(1 + n / i, m + 1);
		if (l >= r)continue;
		ans -= (mint(n + 1))*i*(f(r) - f(l));
		ans += (mint(i))*(i + 1)*(f2(r) - f2(l)) * inv2;
	}
	cout << ans.val() << endl;
	return 0;
}