#include <bits/stdc++.h>
using namespace std;
long long s, x, p, o;
long long ans;
int xo[66] = {}, r[66] = {}, xoo, ro;
int main() {
  cin >> s >> x;
  ans = 1ll;
  o = x;
  p = s - x;
  if (p % 2 || p < 0) ans = 0ll;
  p = p / 2;
  while (p != 0) {
    r[ro] = p % 2;
    p = p / 2;
    ro++;
  }
  while (x != 0) {
    xo[xoo] = x % 2;
    x /= 2;
    xoo++;
  }
  for (int i = 0; i < xoo || i < ro; i++) {
    if (i < xoo && i < ro) {
      if (xo[i] && !r[i]) ans *= 2ll;
      if (xo[i] && r[i]) ans *= 0ll;
    } else {
      if (i < xoo) {
        if (xo[i]) ans *= 2ll;
      }
    }
  }
  if (s == o) ans -= 2ll;
  cout << ans;
  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;
long long s, x, p, o;
long long ans;
int xo[66] = {}, r[66] = {}, xoo, ro;
int main() {
  cin >> s >> x;
  ans = 1ll;
  o = x;
  p = s - x;
  if (p % 2 || p < 0) ans = 0ll;
  p = p / 2;
  while (p != 0) {
    r[ro] = p % 2;
    p = p / 2;
    ro++;
  }
  while (x != 0) {
    xo[xoo] = x % 2;
    x /= 2;
    xoo++;
  }
  for (int i = 0; i < xoo || i < ro; i++) {
    if (i < xoo && i < ro) {
      if (xo[i] && !r[i]) ans *= 2ll;
      if (xo[i] && r[i]) ans *= 0ll;
    } else {
      if (i < xoo) {
        if (xo[i]) ans *= 2ll;
      }
    }
  }
  if (s == o) ans -= 2ll;
  cout << ans;
  return 0;
}

```