#include <bits/stdc++.h>
using namespace std;
long long n, m, a[100], b[100], x, y, dp0[100], dp1[100], ans;
int main() {
  cin >> n >> m;
  if (n == m) ans = 2;
  while (n > 0) {
    a[x++] = n & 1;
    n >>= 1;
  }
  while (m > 0) {
    b[y++] = m & 1;
    m >>= 1;
  }
  if (y > x) return cout << 0, 0;
  if (!a[0] && !b[0])
    dp0[0] = dp1[0] = 1;
  else if (a[0] && b[0])
    dp0[0] = 2;
  else
    return cout << 0, 0;
  for (int i = 1; i < x; ++i)
    if (!a[i] && !b[i])
      dp0[i] = dp1[i] = dp0[i - 1];
    else if (!a[i] && b[i])
      dp1[i] = (dp1[i - 1] << 1);
    else if (a[i] && !b[i])
      dp0[i] = dp1[i] = dp1[i - 1];
    else
      dp0[i] = (dp0[i - 1] << 1);
  cout << max(dp0[x - 1] - ans, 0ll);
}

### 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 n, m, a[100], b[100], x, y, dp0[100], dp1[100], ans;
int main() {
  cin >> n >> m;
  if (n == m) ans = 2;
  while (n > 0) {
    a[x++] = n & 1;
    n >>= 1;
  }
  while (m > 0) {
    b[y++] = m & 1;
    m >>= 1;
  }
  if (y > x) return cout << 0, 0;
  if (!a[0] && !b[0])
    dp0[0] = dp1[0] = 1;
  else if (a[0] && b[0])
    dp0[0] = 2;
  else
    return cout << 0, 0;
  for (int i = 1; i < x; ++i)
    if (!a[i] && !b[i])
      dp0[i] = dp1[i] = dp0[i - 1];
    else if (!a[i] && b[i])
      dp1[i] = (dp1[i - 1] << 1);
    else if (a[i] && !b[i])
      dp0[i] = dp1[i] = dp1[i - 1];
    else
      dp0[i] = (dp0[i - 1] << 1);
  cout << max(dp0[x - 1] - ans, 0ll);
}

```