#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 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; }
#ifndef KWM_T_MATH_MOBIUS_HPP
#define KWM_T_MATH_MOBIUS_HPP

#include <vector>

namespace kwm_t::math {

/**
 * @brief 1 以上 n 以下の Möbius 関数の値を求める
 *
 * μ(1) = 1
 * μ(n) = 0              : n が平方因子を持つ
 * μ(n) = (-1)^k         : 相異なる素因数を k 個持つ
 *
 * 計算量:
 *   O(n log log n)
 *
 * 空間計算量:
 *   O(n)
 *
 * @param n
 *   Möbius 関数を求める最大値
 *
 * 制約 / 注意:
 *   - 戻り値の i 番目が μ(i)
 *   - μ(0) は 0
 *
 * 使用例:
 *   auto mu = kwm_t::math::mobius_table(n);
 *
 * verified:
 *   https://atcoder.jp/contests/awc0150/submissions/78988183
 */
std::vector<int> mobius_table(int n) {
	std::vector<int> mu(n + 1, 1);
	std::vector<bool> composite(n + 1);

	mu[0] = 0;

	for (int p = 2; p <= n; ++p) {
		if (composite[p]) continue;

		for (int i = p; i <= n; i += p) {
			composite[i] = true;
			mu[i] *= -1;
		}

		for (long long i = 1LL * p * p; i <= n; i += 1LL * p * p) {
			mu[i] = 0;
		}
	}

	return mu;
}

} // namespace kwm_t::math
#endif // KWM_T_MATH_MOBIUS_HPP


int main() {
	std::ios::sync_with_stdio(false);
	std::cin.tie(nullptr);
	long long n, m; cin >> n >> m;
	const int lim = 30000007;
	auto mt = kwm_t::math::mobius_table(lim);


	mint inv2 = mint(2).inv();
	auto s = [&](long long x) {
		return mint(x) * (x + 1) * inv2;
	};

	mint ans = 0;
	rep2(i, 1, lim) {
		mint add = (mint)i*mt[i];
		add *= s(n / i);
		add *= s(m / i);
		ans += add;
	}
	cout << ans.val() << endl;
	return 0;
}