#include <bits/stdc++.h>
using namespace std;
int s[43], x[43];
int n(int i) {
  if (i == 1) return 0;
  return 1;
}
int main() {
  long long int sum, xr, ans = 1;
  int i, j, c1 = 0;
  for (i = 0; i < 43; i++) s[i] = x[i] = 0;
  scanf("%lld %lld", &sum, &xr);
  for (i = 42; sum > 0 || xr > 0; i--) {
    s[i] = sum % 2;
    sum /= 2;
    x[i] = xr % 2;
    if (xr % 2 == 1) c1++;
    xr /= 2;
  }
  int gc = 0, c = 0;
  for (i = 42; i > 0; i--) {
    c = gc;
    if (s[i - 1] == x[i - 1])
      gc = 0;
    else
      gc = 1;
    if ((s[i] && n(x[i]) && c) || (n(s[i]) && n(x[i]) && n(c)) ||
        (gc && n(s[i]) && x[i] && c) || (n(gc) && s[i] && x[i] && n(c)))
      continue;
    else {
      printf("0\n");
      return 0;
    }
  }
  for (i = 0; i < c1; i++) ans *= 2;
  for (i = 42; i >= 0; i--) {
    if (s[i] != x[i]) {
      printf("%lld\n", ans);
      return 0;
    }
  }
  printf("%lld\n", ans - 2);
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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 s[43], x[43];
int n(int i) {
  if (i == 1) return 0;
  return 1;
}
int main() {
  long long int sum, xr, ans = 1;
  int i, j, c1 = 0;
  for (i = 0; i < 43; i++) s[i] = x[i] = 0;
  scanf("%lld %lld", &sum, &xr);
  for (i = 42; sum > 0 || xr > 0; i--) {
    s[i] = sum % 2;
    sum /= 2;
    x[i] = xr % 2;
    if (xr % 2 == 1) c1++;
    xr /= 2;
  }
  int gc = 0, c = 0;
  for (i = 42; i > 0; i--) {
    c = gc;
    if (s[i - 1] == x[i - 1])
      gc = 0;
    else
      gc = 1;
    if ((s[i] && n(x[i]) && c) || (n(s[i]) && n(x[i]) && n(c)) ||
        (gc && n(s[i]) && x[i] && c) || (n(gc) && s[i] && x[i] && n(c)))
      continue;
    else {
      printf("0\n");
      return 0;
    }
  }
  for (i = 0; i < c1; i++) ans *= 2;
  for (i = 42; i >= 0; i--) {
    if (s[i] != x[i]) {
      printf("%lld\n", ans);
      return 0;
    }
  }
  printf("%lld\n", ans - 2);
  return 0;
}

```