#include <bits/stdc++.h>
using namespace std;
long long v[2][2][50];
int main() {
  long long s, x, a = 0, p, q;
  cin >> s >> x;
  int k = 50 - 1;
  p = s;
  q = x;
  while (s || x) {
    int i = s & 1, j = x & 1;
    if (i && !j) {
      if (k == 50 - 1) break;
      v[1][0][k] = v[1][1][k] = v[0][1][k + 1] + v[1][1][k + 1];
    } else if (i && j) {
      if (k == 50 - 1)
        v[1][0][k] = 2;
      else
        v[1][0][k] = 2 * (v[0][0][k + 1] + v[1][0][k + 1]);
    } else if (!i && j) {
      if (k == 50 - 1) break;
      v[0][1][k] = 2 * (v[0][1][k + 1] + v[1][1][k + 1]);
    } else {
      if (k == 50 - 1) {
        v[0][0][k] = v[0][1][k] = 1;
      } else
        v[0][0][k] = v[0][1][k] = v[1][0][k + 1] + v[0][0][k + 1];
    }
    s >>= 1;
    x >>= 1;
    if (!s)
      a = v[i][0][k];
    else
      a = v[i][0][k] + v[i][1][k];
    if (!a) break;
    k--;
  }
  if (a && p == q) a -= 2;
  cout << a << "\n";
  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 v[2][2][50];
int main() {
  long long s, x, a = 0, p, q;
  cin >> s >> x;
  int k = 50 - 1;
  p = s;
  q = x;
  while (s || x) {
    int i = s & 1, j = x & 1;
    if (i && !j) {
      if (k == 50 - 1) break;
      v[1][0][k] = v[1][1][k] = v[0][1][k + 1] + v[1][1][k + 1];
    } else if (i && j) {
      if (k == 50 - 1)
        v[1][0][k] = 2;
      else
        v[1][0][k] = 2 * (v[0][0][k + 1] + v[1][0][k + 1]);
    } else if (!i && j) {
      if (k == 50 - 1) break;
      v[0][1][k] = 2 * (v[0][1][k + 1] + v[1][1][k + 1]);
    } else {
      if (k == 50 - 1) {
        v[0][0][k] = v[0][1][k] = 1;
      } else
        v[0][0][k] = v[0][1][k] = v[1][0][k + 1] + v[0][0][k + 1];
    }
    s >>= 1;
    x >>= 1;
    if (!s)
      a = v[i][0][k];
    else
      a = v[i][0][k] + v[i][1][k];
    if (!a) break;
    k--;
  }
  if (a && p == q) a -= 2;
  cout << a << "\n";
  return 0;
}

```