#include <bits/stdc++.h>
using namespace std;
int x[50], s[50];
long long pd[50][2];
long long go(int i, bool carry) {
  if (i >= 50) return !carry;
  long long &ans = pd[i][carry];
  if (~ans) return ans;
  ans = 0;
  if (x[i] == 0) {
    if (!carry && s[i] == 0)
      ans += go(i + 1, 0);
    else if (carry && s[i] == 1)
      ans += go(i + 1, 0);
  }
  if (x[i] == 0) {
    if (!carry && s[i] == 0)
      ans += go(i + 1, 1);
    else if (carry && s[i] == 1)
      ans += go(i + 1, 1);
  }
  if (x[i] == 1) {
    if (!carry && s[i] == 1)
      ans += 2 * go(i + 1, 0);
    else if (carry && s[i] == 0)
      ans += 2 * go(i + 1, 1);
  }
  return ans;
}
int main() {
  long long a, b;
  scanf("%lld%lld", &a, &b);
  for (int(i) = (0); (i) < (50); ++(i))
    s[i] = (a >> i) & 1, x[i] = (b >> i) & 1;
  memset((pd), (-1), sizeof(pd));
  long long ans = go(0, 0);
  if (a == b) ans -= 2;
  printf("%lld\n", ans);
}

### 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 x[50], s[50];
long long pd[50][2];
long long go(int i, bool carry) {
  if (i >= 50) return !carry;
  long long &ans = pd[i][carry];
  if (~ans) return ans;
  ans = 0;
  if (x[i] == 0) {
    if (!carry && s[i] == 0)
      ans += go(i + 1, 0);
    else if (carry && s[i] == 1)
      ans += go(i + 1, 0);
  }
  if (x[i] == 0) {
    if (!carry && s[i] == 0)
      ans += go(i + 1, 1);
    else if (carry && s[i] == 1)
      ans += go(i + 1, 1);
  }
  if (x[i] == 1) {
    if (!carry && s[i] == 1)
      ans += 2 * go(i + 1, 0);
    else if (carry && s[i] == 0)
      ans += 2 * go(i + 1, 1);
  }
  return ans;
}
int main() {
  long long a, b;
  scanf("%lld%lld", &a, &b);
  for (int(i) = (0); (i) < (50); ++(i))
    s[i] = (a >> i) & 1, x[i] = (b >> i) & 1;
  memset((pd), (-1), sizeof(pd));
  long long ans = go(0, 0);
  if (a == b) ans -= 2;
  printf("%lld\n", ans);
}

```