#include <bits/stdc++.h>
using namespace std;
int main() {
  long long s, x;
  cin >> s >> x;
  if (x > s) {
    cout << 0;
    return 0;
  }
  long long s1 = s;
  long long x1 = x;
  long long cp = 1;
  static int b[100];
  int p = 0;
  while (x1 != 0) {
    b[p] = x1 % 2;
    p++;
    s1 -= cp * (long long)(x1 % 2);
    x1 /= 2;
    cp *= 2;
  }
  if (s1 % 2 == 1) {
    cout << 0;
    return 0;
  }
  s1 /= 2;
  static int v[100];
  p = 0;
  while (s1 != 0) {
    v[p] = s1 % 2;
    s1 /= 2;
    p++;
  }
  long long ans = 1;
  for (int i = 0; i < 100; i++) {
    if (b[i] and v[i]) {
      cout << 0;
      return 0;
    }
    if (b[i]) {
      ans *= 2;
    }
  }
  if (s == x) {
    ans -= 2;
  }
  cout << ans;
}

### Prompt

Please formulate a cpp solution 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;
int main() {
  long long s, x;
  cin >> s >> x;
  if (x > s) {
    cout << 0;
    return 0;
  }
  long long s1 = s;
  long long x1 = x;
  long long cp = 1;
  static int b[100];
  int p = 0;
  while (x1 != 0) {
    b[p] = x1 % 2;
    p++;
    s1 -= cp * (long long)(x1 % 2);
    x1 /= 2;
    cp *= 2;
  }
  if (s1 % 2 == 1) {
    cout << 0;
    return 0;
  }
  s1 /= 2;
  static int v[100];
  p = 0;
  while (s1 != 0) {
    v[p] = s1 % 2;
    s1 /= 2;
    p++;
  }
  long long ans = 1;
  for (int i = 0; i < 100; i++) {
    if (b[i] and v[i]) {
      cout << 0;
      return 0;
    }
    if (b[i]) {
      ans *= 2;
    }
  }
  if (s == x) {
    ans -= 2;
  }
  cout << ans;
}

```