#include <bits/stdc++.h>
#pragma warning(disable : 4996)
using namespace std;
long long s, x;
int a[60];
int u[60];
int main() {
  int i, j, k;
  cin >> s >> x;
  if (s % 2 != x % 2) return !printf("0");
  for (i = 1; i < 60; i++) {
    if ((s & (1ll << i)) != (x & (1ll << i))) a[i - 1] = 1;
  }
  long long ans = 1;
  int nowcarry = 0;
  for (i = 0; i < 60; i++) {
    if (x & (1ll << i)) {
      u[i] = 1;
      ans *= 2;
    } else {
      if (a[i] == 1) {
        u[i] = 2;
      } else
        u[i] = 0;
    }
  }
  long long sum = 0;
  for (i = 0; i < 60; i++) sum += (1ll << i) * u[i];
  if (sum != s) return !printf("0");
  if (s == x) ans -= 2;
  cout << ans;
  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>
#pragma warning(disable : 4996)
using namespace std;
long long s, x;
int a[60];
int u[60];
int main() {
  int i, j, k;
  cin >> s >> x;
  if (s % 2 != x % 2) return !printf("0");
  for (i = 1; i < 60; i++) {
    if ((s & (1ll << i)) != (x & (1ll << i))) a[i - 1] = 1;
  }
  long long ans = 1;
  int nowcarry = 0;
  for (i = 0; i < 60; i++) {
    if (x & (1ll << i)) {
      u[i] = 1;
      ans *= 2;
    } else {
      if (a[i] == 1) {
        u[i] = 2;
      } else
        u[i] = 0;
    }
  }
  long long sum = 0;
  for (i = 0; i < 60; i++) sum += (1ll << i) * u[i];
  if (sum != s) return !printf("0");
  if (s == x) ans -= 2;
  cout << ans;
  return 0;
}

```