#include <bits/stdc++.h>
using namespace std;
const int N = 55;
int main() {
  long long int s, x;
  cin >> s >> x;
  long long int ans = 0;
  if (s == x) {
    int qnt = 0;
    for (int i = 0; i < N; i++) {
      if (x & (1ll << i)) {
        qnt++;
      }
    }
    ans = (1ll << qnt) - 2;
  } else {
    int qnt = 0;
    long long int pode = 0;
    for (int i = 0; i < N; i++) {
      if (!(x & (1ll << i))) pode |= (1ll << (i + 1));
    }
    long long int tem = s - x;
    if ((tem & pode) != tem)
      ans = 0;
    else {
      for (int i = 0; i < N; i++) {
        if (x & (1ll << i)) {
          qnt++;
        }
      }
      ans = (1ll << qnt);
    }
  }
  cout << ans << '\n';
  return 0;
}

### Prompt

Create a solution in CPP for 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;
const int N = 55;
int main() {
  long long int s, x;
  cin >> s >> x;
  long long int ans = 0;
  if (s == x) {
    int qnt = 0;
    for (int i = 0; i < N; i++) {
      if (x & (1ll << i)) {
        qnt++;
      }
    }
    ans = (1ll << qnt) - 2;
  } else {
    int qnt = 0;
    long long int pode = 0;
    for (int i = 0; i < N; i++) {
      if (!(x & (1ll << i))) pode |= (1ll << (i + 1));
    }
    long long int tem = s - x;
    if ((tem & pode) != tem)
      ans = 0;
    else {
      for (int i = 0; i < N; i++) {
        if (x & (1ll << i)) {
          qnt++;
        }
      }
      ans = (1ll << qnt);
    }
  }
  cout << ans << '\n';
  return 0;
}

```