#include <bits/stdc++.h>
using namespace std;
long long s, x;
map<long long, bool> mp;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> s >> x;
  long long res = 0LL;
  long long cont = 0LL;
  for (long long i = 0LL; i < 63LL; ++i)
    if (x & (1LL << i)) {
      res += (1LL << i);
      ++cont;
    } else
      mp[1LL << (i + 1LL)] = true;
  if (res <= s) {
    long long dif = s - res;
    bool fl = true;
    for (long long i = 0LL; i < 63LL; ++i) {
      if (dif & (1LL << i)) fl = min(mp[1LL << i], fl);
    }
    if (fl) {
      if (dif == 0LL)
        cout << (1LL << cont) - 2LL << '\n';
      else
        cout << (1LL << cont) << '\n';
    } else
      cout << "0\n";
  } else
    cout << "0\n";
  return 0;
}

### Prompt

Please create a solution in cpp to the following problem:

Two positive integers a and b have a sum of s and a bitwise XOR of x. How many possible values are there for the ordered pair (a, b)?

Input

The first line of the input contains two integers s and x (2 ≤ s ≤ 1012, 0 ≤ x ≤ 1012), the sum and bitwise xor of the pair of positive integers, respectively.

Output

Print a single integer, the number of solutions to the given conditions. If no solutions exist, print 0.

Examples

Input

9 5


Output

4


Input

3 3


Output

2


Input

5 2


Output

0

Note

In the first sample, we have the following solutions: (2, 7), (3, 6), (6, 3), (7, 2).

In the second sample, the only solutions are (1, 2) and (2, 1).

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long s, x;
map<long long, bool> mp;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> s >> x;
  long long res = 0LL;
  long long cont = 0LL;
  for (long long i = 0LL; i < 63LL; ++i)
    if (x & (1LL << i)) {
      res += (1LL << i);
      ++cont;
    } else
      mp[1LL << (i + 1LL)] = true;
  if (res <= s) {
    long long dif = s - res;
    bool fl = true;
    for (long long i = 0LL; i < 63LL; ++i) {
      if (dif & (1LL << i)) fl = min(mp[1LL << i], fl);
    }
    if (fl) {
      if (dif == 0LL)
        cout << (1LL << cont) - 2LL << '\n';
      else
        cout << (1LL << cont) << '\n';
    } else
      cout << "0\n";
  } else
    cout << "0\n";
  return 0;
}

```