ABC 227

using Int = unsigned long long int;

Int kth_root(Int x, Int k) {
    assert(k != 0);
    if (x == 1 || k == 1) return x;
    Int l = 0, r = x;
    while (r - l > 1) {
        Int m = (r - l) / 2 + l;
        Int t = x;
        rep(i, k) t /= m;
        if (1 > t) {
            r = m;
        } else {
            l = m;
        }
    }
    return l;
};

// 平方根
Int isqrt(Int x) {
    return kth_root(x, 2);
};

vll cal_lpf(ll N) {
    vll lpf(N + 1, -1);
    vll primes;
    rep2(d, 2, N + 1) {
        if (lpf[d] < 0) {
            lpf[d] = d;
            primes.push_back(d);
        }
        for (ll p : primes) {
            if (p * d > N || p > lpf[d]) break;
            lpf[p * d] = p;
        }
    }

    return lpf;
}

// x を素因数分解する
// pair = <prime, cnt>
vector<pair<ll, ll>> factorize(vll& lpf, ll x) {
    vector<pair<ll, ll>> facs;
    {
        while (x != 1) {
            ll p = lpf[x];
            ll cnt = 0;
            while (x % p == 0) {
                x /= p;
                cnt++;
            }
            facs.push_back({p, cnt});
        }
    }

    return facs;
}

void solve() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

    ll N, K;
    cin >> N >> K;

    ll M = max(K + 1, (ll)isqrt(N) + 1);
    vll lpf = cal_lpf(M), primes;
    rep2(i, 2, M) {
        if (lpf[i] == i) primes.push_back(i);
    }

    ll psz = primes.size();
    vll cnt(psz);
    rep2(k, 2, K + 1) {
        auto ps = factorize(lpf, k);
        for (auto [p, num] : ps) {
            int i = lower_bound(all(primes), p) - primes.begin();
            cnt[i] -= num;
        }
    }

    ll L = N - K + 1, R = N;
    vll A(R - L + 1);
    iota(all(A), L);
    for (ll p : primes) {
        for (ll j = ceil(L, p) * p; j <= R; j += p) {
            ll x = A[j - L];
            ll num = 0;
            while (x % p == 0) {
                num++;
                x /= p;
            }
            int i = lower_bound(all(primes), p) - primes.begin();
            cnt[i] += num;
            A[j - L] = x;
        }
    }

    mint ans = 1;
    for (ll x : cnt) ans *= x + 1;
    for (ll x : A) {
        if (x != 1) ans *= 2;
    }

    cout << ans.val() << endl;
}