#include <bits/stdc++.h>
using namespace std;
long long sum[3000005];
int main() {
  long long s, x;
  cin >> s >> x;
  long long l1 = log(s) / log(2) / 2 + 1;
  long long a = 0;
  for (long long i = 0; i < (1ll << l1); i++) {
    a = i ^ (x % (1ll << l1));
    sum[i + a]++;
  }
  long long l2 = log(s) / log(2) + 1 - l1;
  long long ans = 0;
  for (long long i = 0; i < (1ll << l2); i++) {
    a = i ^ (x / (1ll << l1));
    if (s >= (i + a) * (1ll << l1) && (s - (i + a) * (1ll << l1) < 3000000))
      ans += sum[s - (i + a) * (1ll << l1)];
  }
  if (0ll ^ s == x) ans -= 2;
  cout << ans;
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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 sum[3000005];
int main() {
  long long s, x;
  cin >> s >> x;
  long long l1 = log(s) / log(2) / 2 + 1;
  long long a = 0;
  for (long long i = 0; i < (1ll << l1); i++) {
    a = i ^ (x % (1ll << l1));
    sum[i + a]++;
  }
  long long l2 = log(s) / log(2) + 1 - l1;
  long long ans = 0;
  for (long long i = 0; i < (1ll << l2); i++) {
    a = i ^ (x / (1ll << l1));
    if (s >= (i + a) * (1ll << l1) && (s - (i + a) * (1ll << l1) < 3000000))
      ans += sum[s - (i + a) * (1ll << l1)];
  }
  if (0ll ^ s == x) ans -= 2;
  cout << ans;
  return 0;
}

```