如果这三个数相同，那么对答案贡献为0，如果不都相同，对答案贡献为22^i，直接拿r和l比较，从高位开始，如果出现r的二进制位与l的二进制位不同，那么r的必定是1，l的必定是0，因为r>l，那么之后的所有二进制位r全部设为0，l全部设为1，分别得到a，b，这样才能保证设好的这两个数a,b一定在lr区间内，至于第三个数，可以是a+1，或者b-1，这样才能保证每一位的贡献是22^i

const int N = 1e6 + 10;


string bin2(int n) {
    string str = "";
    while (n != 0) {
        str = to_string(n % 2) + str;
        n = n / 2;
    }
    return str;
}
int bin10(string n)
{

    int sum = 0;
    for (int i = 0; i < n.size(); i++)
    {
        if (n[i] == '1')
        {
            int j = pow(2, n.size() - i - 1);
            sum += j;
        }
    }
    return sum;
}
void solve()
{
    int l, r; cin >> l >> r;
    string sl = bin2(l);
    string sr = bin2(r);
    while (sr.size() > sl.size()) sl = "0" + sl;
    int n = sr.size();
    sl = " " + sl;
    sr = " " + sr;
    int len = 0;
    int ok = 0;
    for (int i = 1; i <= sr.size(); i++)
    {
        if (sl[i] != sr[i])
        {
            ok = i;
            break;
        }
    }
    string a, b;
    for (int i = 1; i <= n; i++)
    {
        if (i > ok)
        {
            a += "0";
            b += "1";
        }
        else
        {
            a += sr[i];
            b += sl[i];
        }
    }
    int oa = bin10(a);
    int ob = bin10(b);
    if (oa + 1 <= r) cout << oa + 1 << " " << oa << " " << ob << '\n';
    else cout << oa << " " << ob << " " << ob - 1 << '\n';

}