#include <bits/stdc++.h>
using namespace std;
long long x, y;
int a[55], b[55];
int main() {
  cin >> x >> y;
  int cnt = 0;
  for (int i = 0; i < 50; i++) {
    if (x & (1LL << i)) {
      a[i] = 1;
    }
    if (y & (1LL << i)) {
      b[i] = 1;
      cnt++;
    }
  }
  for (int i = 49; i >= 0; i--) {
    if (!a[i] && b[i]) {
      if (!(a[i + 1] ^ b[i + 1])) {
        cout << "0" << endl;
        return 0;
      }
    }
    if (a[i] ^ b[i]) {
      if (!i || (b[i - 1] && a[i - 1])) {
        cout << "0" << endl;
        return 0;
      }
    }
  }
  long long ret = 1;
  for (int i = 1; i <= cnt; i++) ret <<= 1;
  if (y == x) ret -= 2;
  cout << ret << endl;
  return 0;
}

### Prompt

Generate 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;
long long x, y;
int a[55], b[55];
int main() {
  cin >> x >> y;
  int cnt = 0;
  for (int i = 0; i < 50; i++) {
    if (x & (1LL << i)) {
      a[i] = 1;
    }
    if (y & (1LL << i)) {
      b[i] = 1;
      cnt++;
    }
  }
  for (int i = 49; i >= 0; i--) {
    if (!a[i] && b[i]) {
      if (!(a[i + 1] ^ b[i + 1])) {
        cout << "0" << endl;
        return 0;
      }
    }
    if (a[i] ^ b[i]) {
      if (!i || (b[i - 1] && a[i - 1])) {
        cout << "0" << endl;
        return 0;
      }
    }
  }
  long long ret = 1;
  for (int i = 1; i <= cnt; i++) ret <<= 1;
  if (y == x) ret -= 2;
  cout << ret << endl;
  return 0;
}

```