#include <bits/stdc++.h>
using namespace std;
int b[1004];
int main() {
  long long s, x;
  cin >> s >> x;
  long long y = x;
  int z = 0;
  for (int i = 0; i <= 40; i++) {
    z += y % 2;
    y = y / 2LL;
  }
  long long ans = 1;
  for (int i = 1; i <= z; i++) ans = ans * 2LL;
  if (s == x) {
    cout << ans - 2LL << '\n';
    return 0;
  }
  if (s < x) {
    cout << 0 << '\n';
    return 0;
  }
  long long t = s - x;
  if (t % 2 != 0) {
    cout << 0 << '\n';
    return 0;
  }
  t = t / 2LL;
  for (int i = 0; i <= 40; i++) {
    if (x % 2 == 1 && t % 2 == 1) {
      cout << 0 << '\n';
      return 0;
    }
    x = x / 2LL;
    t = t / 2LL;
  }
  cout << ans << '\n';
}

### Prompt

Your task is to create 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 b[1004];
int main() {
  long long s, x;
  cin >> s >> x;
  long long y = x;
  int z = 0;
  for (int i = 0; i <= 40; i++) {
    z += y % 2;
    y = y / 2LL;
  }
  long long ans = 1;
  for (int i = 1; i <= z; i++) ans = ans * 2LL;
  if (s == x) {
    cout << ans - 2LL << '\n';
    return 0;
  }
  if (s < x) {
    cout << 0 << '\n';
    return 0;
  }
  long long t = s - x;
  if (t % 2 != 0) {
    cout << 0 << '\n';
    return 0;
  }
  t = t / 2LL;
  for (int i = 0; i <= 40; i++) {
    if (x % 2 == 1 && t % 2 == 1) {
      cout << 0 << '\n';
      return 0;
    }
    x = x / 2LL;
    t = t / 2LL;
  }
  cout << ans << '\n';
}

```