#include <bits/stdc++.h>
using namespace std;
int main() {
  long long x, s;
  while (cin >> s >> x) {
    long long ans = 1, l = -1;
    bool inv = false;
    int pp = 0;
    for (int i = int(0); i < int(60); i++) {
      long long j = 1LL << (long long)i;
      if (x & j) {
        if (s & j) {
          if (pp == 2) inv = true;
          pp = 0;
          ans *= 2;
        } else {
          if (!pp) inv = true;
          ans *= 2;
          pp = 2;
        }
      } else {
        if (s & j) {
          if (!pp) inv = true;
          pp = 1;
        } else {
          if (pp == 2) inv = true;
          pp = 1;
        }
      }
    }
    if (s == x) ans -= 2;
    if (inv) ans = 0;
    cout << ans << endl;
  }
  return 0;
}

### Prompt

Please create a solution in Cpp 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;
int main() {
  long long x, s;
  while (cin >> s >> x) {
    long long ans = 1, l = -1;
    bool inv = false;
    int pp = 0;
    for (int i = int(0); i < int(60); i++) {
      long long j = 1LL << (long long)i;
      if (x & j) {
        if (s & j) {
          if (pp == 2) inv = true;
          pp = 0;
          ans *= 2;
        } else {
          if (!pp) inv = true;
          ans *= 2;
          pp = 2;
        }
      } else {
        if (s & j) {
          if (!pp) inv = true;
          pp = 1;
        } else {
          if (pp == 2) inv = true;
          pp = 1;
        }
      }
    }
    if (s == x) ans -= 2;
    if (inv) ans = 0;
    cout << ans << endl;
  }
  return 0;
}

```