#include <bits/stdc++.h>
using namespace std;
const int MAXN = 2e5 + 5;
long long s, x;
int same;
bool ok[65];
long long ct = 1;
int main() {
  cin >> s >> x;
  same = (s == x);
  for (long long i = 0; i < 62; ++i) {
    ok[i] = x % 2;
    if (ok[i] == 1) ct *= 2;
    if (ok[i] == 1) s -= ((1LL) << i);
    x /= 2;
  }
  if (s < 0) {
    cout << "0\n";
    return 0;
  }
  for (int i = 0; i < 62; ++i) {
    if (s % 2 == 0) {
      s /= 2;
      continue;
    }
    if (i == 0) {
      cout << "0\n";
      return 0;
    }
    if (i > 0 && ok[i - 1]) {
      cout << "0\n";
      return 0;
    }
    s /= 2;
  }
  cout << ct - 2 * same << 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;
const int MAXN = 2e5 + 5;
long long s, x;
int same;
bool ok[65];
long long ct = 1;
int main() {
  cin >> s >> x;
  same = (s == x);
  for (long long i = 0; i < 62; ++i) {
    ok[i] = x % 2;
    if (ok[i] == 1) ct *= 2;
    if (ok[i] == 1) s -= ((1LL) << i);
    x /= 2;
  }
  if (s < 0) {
    cout << "0\n";
    return 0;
  }
  for (int i = 0; i < 62; ++i) {
    if (s % 2 == 0) {
      s /= 2;
      continue;
    }
    if (i == 0) {
      cout << "0\n";
      return 0;
    }
    if (i > 0 && ok[i - 1]) {
      cout << "0\n";
      return 0;
    }
    s /= 2;
  }
  cout << ct - 2 * same << endl;
}

```