#include <bits/stdc++.h>
using namespace std;
const int N = 64;
long long s, x;
long long f[N][2];
void solve() {
  int n = 0, m = 0;
  for (long long t = s; t; t >>= 1) ++n;
  for (long long t = x; t; t >>= 1) ++m;
  n = max(n, m);
  f[0][0] = 1;
  for (int i = 1; i <= n; ++i) {
    int a = (s >> (i - 1)) & 1;
    int b = (x >> (i - 1)) & 1;
    if (!a && b)
      f[i][0] += 0;
    else
      f[i][0] += (b ? 2 : 1) * f[i - 1][a ^ b];
    if (a && b)
      f[i][1] += 0;
    else
      f[i][1] += (b ? 2 : 1) * f[i - 1][a ^ b];
  }
  long long ans = f[n][0];
  if ((0 ^ s) == x) ans -= 2;
  cout << max(0ll, ans) << endl;
}
int main() {
  cin >> s >> x;
  solve();
  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;
const int N = 64;
long long s, x;
long long f[N][2];
void solve() {
  int n = 0, m = 0;
  for (long long t = s; t; t >>= 1) ++n;
  for (long long t = x; t; t >>= 1) ++m;
  n = max(n, m);
  f[0][0] = 1;
  for (int i = 1; i <= n; ++i) {
    int a = (s >> (i - 1)) & 1;
    int b = (x >> (i - 1)) & 1;
    if (!a && b)
      f[i][0] += 0;
    else
      f[i][0] += (b ? 2 : 1) * f[i - 1][a ^ b];
    if (a && b)
      f[i][1] += 0;
    else
      f[i][1] += (b ? 2 : 1) * f[i - 1][a ^ b];
  }
  long long ans = f[n][0];
  if ((0 ^ s) == x) ans -= 2;
  cout << max(0ll, ans) << endl;
}
int main() {
  cin >> s >> x;
  solve();
  return 0;
}

```