The following rules define a kind of integer tuple - the Legend Tuple:
• (1; k) is always a Legend Tuple, where k is an integer.
• if (n; k) is a Legend Tuple, (n + k; k) is also a Legend Tuple.
• if (n; k) is a Legend Tuple, (nk; k) is also a Legend Tuple.
We want to know the number of the Legend Tuples (n; k) where 1 ≤ n ≤ N; 1 ≤ k ≤ K.

很容易想到列舉每個k計算貢獻。

當時畫的草稿：

發現對於每個k。1總是算的，然後就是k,k+1。然後以此為週期。

因此對於一個n，我可以算單個k的貢獻就是 1 + (n - 1) / k 。當然還要加上餘數的部分。

因此總的貢獻就是

當然實現還有很多細節，由於整除分塊寫的不多，很多細節要注意。比如取min，比如溢位，比如先乘還是先除。

int readint() {
    int x = 0, f = 1; char ch = getchar();
    while (ch < '0' || ch>'9') { if (ch == '-')f = -1; ch = getchar(); }
    while (ch >= '0' && ch <= '9') { x = x * 10 + ch - '0'; ch = getchar(); }
    return
 
 x * f;
}

ll readll() {
    ll x = 0, f = 1; char ch = getchar();
    while (ch < '0' || ch>'9') { if (ch == '-')f = -1; ch = getchar(); }
    while (ch >= '0' && ch <= '9') { x = x * 10 + ch - '0'; ch = getchar(); }
    return x * f;
}

void Put(ll x) {
    if (x > 9) Put(x / 10);
    putchar(x % 10 + '0');
}

int main() {
    ll n, k;
    n = readll(), k = readll();
    if (k == 1ll) {
        Put(n % MOD);
        return 0;
    }
    if (n == 1ll) {
        Put(k % MOD);
        return 0;
    }
    ll res = 0;
    for (ll l = 2, r; l <= k && l <= n - 1; l = r + 1) {
        assert(l <= n - 1);
        if (k / l != 0) r = min((n - 1) / ((n - 1) / l), k), r = min(r, n - 1);
        else r = k;
        res = (res + (n - 1) / l * (r - l + 1) % MOD) % MOD;
    }
    res = res * 2 % MOD;
    res = (res + (k % MOD) - 1 + MOD) % MOD;
    for (ll i = 1; i * i <= n; i++) {
        if (n % i) continue;
        if (i <= k) res = (res + 1) % MOD;
        if (n / i <= k && i * i != n) res = (res + 1) % MOD;
    }
    res = (res - 1 + MOD) % MOD;
    res = (res + n) % MOD; 
    Put(res);
}