#include <bits/stdc++.h>
using namespace std;
int a[55];
int main() {
  long long s, x;
  cin >> s >> x;
  int f = 0;
  for (int i = 0; i < 50; i++) {
    if ((x & ((long long)1 << i))) {
      s -= (long long)1 << i;
      a[i] = 1;
      f++;
    }
  }
  if (s % 2 == 1 || s < 0) {
    printf("0\n");
    return 0;
  }
  s = s / 2;
  for (int i = 0; i < 50; i++) {
    if ((s & ((long long)1 << i))) {
      if (a[i]) {
        printf("0\n");
        return 0;
      }
    }
  }
  if (s != 0) {
    long long p = 1;
    for (int i = 1; i <= f; i++) {
      p = p * 2;
    }
    cout << p << endl;
  } else {
    long long p = 1;
    for (int i = 1; i <= f; i++) {
      p = p * 2;
    }
    cout << p - 2 << endl;
  }
}

### Prompt

Please provide a cpp coded solution 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;
int a[55];
int main() {
  long long s, x;
  cin >> s >> x;
  int f = 0;
  for (int i = 0; i < 50; i++) {
    if ((x & ((long long)1 << i))) {
      s -= (long long)1 << i;
      a[i] = 1;
      f++;
    }
  }
  if (s % 2 == 1 || s < 0) {
    printf("0\n");
    return 0;
  }
  s = s / 2;
  for (int i = 0; i < 50; i++) {
    if ((s & ((long long)1 << i))) {
      if (a[i]) {
        printf("0\n");
        return 0;
      }
    }
  }
  if (s != 0) {
    long long p = 1;
    for (int i = 1; i <= f; i++) {
      p = p * 2;
    }
    cout << p << endl;
  } else {
    long long p = 1;
    for (int i = 1; i <= f; i++) {
      p = p * 2;
    }
    cout << p - 2 << endl;
  }
}

```