#include <algorithm>
#include <atcoder/all>
#include <bitset>
#include <cassert>
#include <cmath>
#include <ctime>
#include <functional>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <random>
#include <set>
#include <stack>
#include <string>
#include <tuple>
#include <unordered_map>
#include <unordered_set>
#include <utility>
#include <vector>
using namespace std;
using namespace atcoder;
typedef long long ll;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define repr(i, n) for (int i = (int)(n) - 1; i >= 0; i--)
#define repk(i, k, n) for (int i = k; i < (int)(n); i++)
#define all(v) v.begin(), v.end()
#define mod1 1000000007
#define mod2 998244353
#define mod3 100000007
#define vi vector<int>
#define vs vector<string>
#define vc vector<char>
#define vl vector<ll>
#define vb vector<bool>
#define vvi vector<vector<int>>
#define vvc vector<vector<char>>
#define vvl vector<vector<ll>>
#define vvb vector<vector<bool>>
#define vvvi vector<vector<vector<int>>>
#define vvvl vector<vector<vector<ll>>>
#define pii pair<int, int>
#define pil pair<int, ll>
#define pli pair<ll, int>
#define pll pair<ll, ll>
#define vpii vector<pair<int, int>>
#define vpll vector<pair<ll, ll>>
#define vvpii vector<vector<pair<int, int>>>
#define vvpll vector<vector<pair<ll, ll>>>

template <typename T>
void debug(T e) {
    cerr << e << endl;
}

template <typename T>
void debug(vector<T> &v) {
    rep(i, v.size()) { cerr << v[i] << " "; }
    cerr << endl;
}

template <typename T>
void debug(vector<vector<T>> &v) {
    rep(i, v.size()) {
        rep(j, v[i].size()) { cerr << v[i][j] << " "; }
        cerr << endl;
    }
}

template <typename T>
void debug(vector<pair<T, T>> &v) {
    rep(i, v.size()) { cerr << v[i].first << " " << v[i].second << endl; }
}

template <typename T>
void debug(set<T> &st) {
    for (auto itr = st.begin(); itr != st.end(); itr++) {
        cerr << *itr << " ";
    }
    cerr << endl;
}

template <typename T>
void debug(multiset<T> &ms) {
    for (auto itr = ms.begin(); itr != ms.end(); itr++) {
        cerr << *itr << " ";
    }
    cerr << endl;
}

template <typename T>
void debug(map<T, T> &mp) {
    for (auto itr = mp.begin(); itr != mp.end(); itr++) {
        cerr << itr->first << " " << itr->second << endl;
    }
}

void debug_out() { cerr << endl; }

template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
    cerr << H << " ";
    debug_out(T...);
}

using mint = modint998244353;

void debug_mint1(vector<mint> &vec) {
    for (int i = 0; i < vec.size(); i++) {
        cerr << vec[i].val() << " ";
    }
    cerr << endl;
}

void debug_mint2(vector<vector<mint>> &vec) {
    for (int i = 0; i < vec.size(); i++) {
        for (int j = 0; j < vec[i].size(); j++) {
            cerr << vec[i][j].val() << " ";
        }
        cerr << endl;
    }
}

int main() {
    ll N, M;
    cin >> N >> M;

    // 前問で、\sum (a|x, b|x) μ(x) = 1 であることと、gcd(a, b) = 1 が同値であることを知った
    // これをもとにした式変形ができればよい？
    ll L = max(N, M);

    // mu の和を計算するのに、別の方法はあるか？
    vector<bool> twice(L + 1, false);
    for (ll i = 2; i * i <= L; i++){
        for (ll j = 1; j * i * i <= L; j++){
            twice[j * i * i] = true;
        }
    }
    vector<bool> isprime(L + 1, true);
    isprime[0] = false;
    isprime[1] = false;
    for (ll i = 2; i <= L; i++){
        if (!isprime[i]) continue;
        for (ll j = 2; j * i <= L; j++){
            isprime[i * j] = false;
        }
    }
    
    vector<ll> cnts(L + 1);
    for (ll i = 2; i <= L; i++){
        if (isprime[i]){
            for (ll j = 1; j * i <= L; j++){
                cnts[i * j]++;
            }
        }
    }
    vector<ll> mu(L + 1, 0);
    for (ll i = 1; i <= L; i++){
        if (!twice[i]){
            if (cnts[i] % 2 == 0) mu[i] = 1;
            else mu[i] = -1;
        }
    }

    mint ans = mint(0);
    mint inv_4 = mint(1) / mint(4);
    for (ll d = 1; d <= L; d++){
        mint max_n = mint(N / d);
        mint max_m = mint(M / d);
        ans += mint(mu[d]) * mint(d) * mint(d) * (max_n * (max_n + 1)) * (max_m * (max_m + 1)) * inv_4;
    }
    cout << ans.val() << endl;
}