#include <bits/stdc++.h>
using namespace std;
long long a[100], b[100], c[100], ans = 1, sum;
int main() {
  long long s, x, ss;
  cin >> s >> x;
  ss = s;
  for (long long i = 0; i <= 50ll; i++) {
    if ((long long)(1ll << i) & x) {
      sum += (long long)(1ll << i);
      ans *= (long long)2ll;
    } else {
      b[i] = 1;
    }
  }
  s -= sum;
  if (s < 0) {
    cout << 0 << endl;
    return 0;
  }
  for (long long i = 0; i <= 50ll; i++) {
    if ((long long)(1ll << i) & s) {
      c[i] = 1;
    }
  }
  if (c[0] == 1) {
    cout << 0 << endl;
    return 0;
  }
  for (int i = 1; i <= 50; i++) {
    if (c[i] == 1) {
      if (b[i - 1] == 0) {
        cout << 0 << endl;
        return 0;
      }
    }
  }
  if (ss == x) {
    ans -= 2;
  }
  cout << ans << endl;
}

### Prompt

In cpp, your task is to solve 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 a[100], b[100], c[100], ans = 1, sum;
int main() {
  long long s, x, ss;
  cin >> s >> x;
  ss = s;
  for (long long i = 0; i <= 50ll; i++) {
    if ((long long)(1ll << i) & x) {
      sum += (long long)(1ll << i);
      ans *= (long long)2ll;
    } else {
      b[i] = 1;
    }
  }
  s -= sum;
  if (s < 0) {
    cout << 0 << endl;
    return 0;
  }
  for (long long i = 0; i <= 50ll; i++) {
    if ((long long)(1ll << i) & s) {
      c[i] = 1;
    }
  }
  if (c[0] == 1) {
    cout << 0 << endl;
    return 0;
  }
  for (int i = 1; i <= 50; i++) {
    if (c[i] == 1) {
      if (b[i - 1] == 0) {
        cout << 0 << endl;
        return 0;
      }
    }
  }
  if (ss == x) {
    ans -= 2;
  }
  cout << ans << endl;
}

```