public static long sumXor(long n) {
        // Write your code here
        if(n == 0) return 1 ;
        
        int totalOneBits = Long.bitCount(n) ;
        int totalBits = Long.toBinaryString(n).length() ;
         int zeroBits = totalBits - totalOneBits ;
         
         return 1l << zeroBits ; //bitwise left shift => 8 4 2 1 ;power of 2
    }

public static long sumXor(long n) {
        List<Long> xorsList = new ArrayList<>();
        xorsList.add(0l);
        for(int i=0;i<64;i++){
            long powerOfTwo = (long)Math.pow(2, i);
            if(powerOfTwo>n)
                break;
            if((powerOfTwo+n)==(powerOfTwo^n)){
                xorsList.add(powerOfTwo);
            }
        }
        if(xorsList.size()==1)
            return xorsList.size();
        Set<Long> resultList = new HashSet<>(xorsList);
        resultList = scrambleXORS(xorsList,resultList,n,xorsList.get(0),0);
        return resultList.size();
    }
    public static Set<Long> scrambleXORS(List<Long> baseXORS,Set<Long> resultList,
                                          long n, long currNumber,int index) {
        if(index>=baseXORS.size()){
            resultList.add(currNumber);
            return resultList;
        }
        for(int i=index;i<baseXORS.size();i++)
        {
            Long xorBase = baseXORS.get(i);
            long currSum = currNumber+xorBase;
            resultList.add(currSum);
            if(currSum<n)
            {
                scrambleXORS(baseXORS,resultList,n,currSum, i+1);
            }
        }
        return resultList;
    }

fn sumXor(n: i64) -> i64 {
    2i64.pow(n.count_zeros() - n.leading_zeros())
}

long sumXor(long n) {
    if (n == 0) {
        return 1;
    }
    
    int num_bits = std::floor(std::log2(n));
    int num_one_bits = __builtin_popcountl(n) - 1; // -1 excludes highest power of 2
    int num_zero_bits = num_bits - num_one_bits;
    
    // # of ways of choosing num_zero_bits
    return 1L << num_zero_bits; // pow(2, num_zero_bits)
}

def sumXor(n):
    count = 1
    while n != 0:
        if n % 2 == 0:
            count *= 2
        n >>= 1
    return count