#include <bits/stdc++.h>

#define int long long

const int p = 998244353;

const int N = 1e5 + 9;
int fac[N], ifac[N], inv[N];
int qmi(int a, int b)
{
    int res = 1;
    while (b)
    {
        if (b & 1)

        {
            res = res * a % p;
        }
        a = a * a % p;
        b >>= 1;
    }
    return res;
}
int INV(int a)
{
    return qmi(a, p - 2);
}

int C(int a, int b)
{
    if (a < b or std::min(a, b) < 0)
    {
        return 0;
    }
    return fac[a] * ifac[a - b] % p * ifac[b] % p;
}
signed main()
{
    std::ios::sync_with_stdio(0);
    std::cin.tie(0);

    int m, n;
    std::cin >> m >> n;

    fac[0] = 1;
    for (int i = 1; i <= m; ++i)
    {
        fac[i] = fac[i - 1] * i % p;
    }
    ifac[m] = INV(fac[m]);
    for (int i = m; i >= 1; --i)
    {
        ifac[i - 1] = i * ifac[i] % p;
        inv[i] = ifac[i] * fac[i - 1] % p;
    }

    std::vector dp(21, std::vector<int>(m + 1, 0));
    for (int i = 1; i <= m; ++i)
    {
        dp[0][i] = 1;
    }
    for (int i = 1; i <= m; ++i)
    {
        for (int j = 2; j * i <= m; ++j)
        {
            int now = j * i;
            for (int time = 1; time <= 20; ++time)
            {
                dp[time][now] = (dp[time][now] + dp[time - 1][i]) % p;
            }
        }
    }

    int ans = C(m, n) * n % p;
    for (int last = 1; last <= m; ++last)
    {
        for (int val = 2 * last; val <= m; val += last)
        {
            int cnt2 = m / val - 1; // 是val的倍数
            int cnt1 = m / last - val / last - cnt2;
            for (int num = n - 22; num <= n - 2; ++num)
            {
                int pre = n - num - 2; // 前面有多少个
                ans = (ans - dp[pre][last] * ((C(cnt1 + cnt2, num) - C(cnt2, num) + p) % p) % p * (pre + 1) % p + p) % p;
            }
        }
    }

    for (int i = 1; i <= m; ++i)
    {
        if (n - 1 <= 20)
        {
            ans = (ans - n * dp[n - 1][i] % p + p) % p;
        }
    }

    std::cout << ans << '\n';
}