#include <bits/stdc++.h>
using namespace std;
int main() {
  long long s, x;
  cin >> s >> x;
  long long d = s - x;
  if (d % 2 || x > s) {
    cout << 0 << endl;
    return 0;
  }
  int c = 0;
  vector<bool> bt(60);
  int t = 0;
  while (x > 0) {
    if (x % 2) {
      c++;
      bt[t] = true;
    }
    t++;
    x /= 2;
  }
  long long r = d / 2;
  t = 0;
  while (r > 0) {
    if (r % 2 && bt[t]) {
      cout << 0 << endl;
      return 0;
    }
    t++;
    r /= 2;
  }
  long long anw = 1;
  for (int i = 0; i < c; i++) anw *= 2;
  if (d == 0) anw -= 2;
  cout << anw << endl;
}

### 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 main() {
  long long s, x;
  cin >> s >> x;
  long long d = s - x;
  if (d % 2 || x > s) {
    cout << 0 << endl;
    return 0;
  }
  int c = 0;
  vector<bool> bt(60);
  int t = 0;
  while (x > 0) {
    if (x % 2) {
      c++;
      bt[t] = true;
    }
    t++;
    x /= 2;
  }
  long long r = d / 2;
  t = 0;
  while (r > 0) {
    if (r % 2 && bt[t]) {
      cout << 0 << endl;
      return 0;
    }
    t++;
    r /= 2;
  }
  long long anw = 1;
  for (int i = 0; i < c; i++) anw *= 2;
  if (d == 0) anw -= 2;
  cout << anw << endl;
}

```