#include <bits/stdc++.h>
using namespace std;
bool a[40], b[40];
int bit(long long a, int p) { return (bool)(a & ((long long)1 << p)); }
void st(long long a, bool s[40]) {
  for (int i = 0; i <= 40; i++) s[i] = bit(a, i);
}
long long f(int pos, int fl) {
  if (pos == 40) return 1 - fl;
  int q = a[pos] - fl;
  if (b[pos]) {
    if (q == 1) return 2 * f(pos + 1, 0);
    if (q == 0) return 0;
    if (q == -1) return 2 * f(pos + 1, 1);
  } else {
    if (q == 0)
      return f(pos + 1, 0) + f(pos + 1, 1);
    else
      return 0;
  }
}
int main() {
  long long x, y;
  cin >> x >> y;
  st(x, a);
  st(y, b);
  long long w = f(0, 0);
  if (x == y) w -= 2;
  cout << w << "\n";
  return 0;
}

### Prompt

Your challenge is to write 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;
bool a[40], b[40];
int bit(long long a, int p) { return (bool)(a & ((long long)1 << p)); }
void st(long long a, bool s[40]) {
  for (int i = 0; i <= 40; i++) s[i] = bit(a, i);
}
long long f(int pos, int fl) {
  if (pos == 40) return 1 - fl;
  int q = a[pos] - fl;
  if (b[pos]) {
    if (q == 1) return 2 * f(pos + 1, 0);
    if (q == 0) return 0;
    if (q == -1) return 2 * f(pos + 1, 1);
  } else {
    if (q == 0)
      return f(pos + 1, 0) + f(pos + 1, 1);
    else
      return 0;
  }
}
int main() {
  long long x, y;
  cin >> x >> y;
  st(x, a);
  st(y, b);
  long long w = f(0, 0);
  if (x == y) w -= 2;
  cout << w << "\n";
  return 0;
}

```