#include <bits/stdc++.h>
using namespace std;
long long s, x;
int main() {
  scanf("%lld%lld", &s, &x);
  long long ans = 0ll, carry0 = 1ll, carry1 = 0ll, i = 0;
  long long m = max(s, x);
  long long bi = 0;
  while (m != 0) {
    bi++;
    m = m / 2;
  }
  for (i = 0; i < bi; i++) {
    long long ac0 = 0ll, ac1 = 0ll;
    if (((x & (1ll << i)) == 0ll) && ((s & (1ll << i)) == 0ll)) {
      ac1 = carry0;
      carry1 = 0ll;
    } else if (((x & (1ll << i)) != 0ll) && ((s & (1ll << i)) == 0ll)) {
      carry0 = 0ll;
      ac1 = carry1;
    } else if (((x & (1ll << i)) == 0ll) && ((s & (1ll << i)) != 0ll)) {
      carry0 = 0ll;
      ac0 = carry1;
    } else {
      carry1 = 0ll;
      ac0 = carry0;
    }
    carry0 += ac0;
    carry1 += ac1;
  }
  if (s == x) carry0 -= 2;
  printf("%lld\n", carry0);
  return 0;
}

### Prompt

Create a solution in CPP for 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;
int main() {
  scanf("%lld%lld", &s, &x);
  long long ans = 0ll, carry0 = 1ll, carry1 = 0ll, i = 0;
  long long m = max(s, x);
  long long bi = 0;
  while (m != 0) {
    bi++;
    m = m / 2;
  }
  for (i = 0; i < bi; i++) {
    long long ac0 = 0ll, ac1 = 0ll;
    if (((x & (1ll << i)) == 0ll) && ((s & (1ll << i)) == 0ll)) {
      ac1 = carry0;
      carry1 = 0ll;
    } else if (((x & (1ll << i)) != 0ll) && ((s & (1ll << i)) == 0ll)) {
      carry0 = 0ll;
      ac1 = carry1;
    } else if (((x & (1ll << i)) == 0ll) && ((s & (1ll << i)) != 0ll)) {
      carry0 = 0ll;
      ac0 = carry1;
    } else {
      carry1 = 0ll;
      ac0 = carry0;
    }
    carry0 += ac0;
    carry1 += ac1;
  }
  if (s == x) carry0 -= 2;
  printf("%lld\n", carry0);
  return 0;
}

```