#include <bits/stdc++.h>
using namespace std;
int A[64], B[64];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long s, x, i, j, k, ans = 1, carry, extra = 2;
  cin >> s >> x;
  for (i = 0, j = 1; i < 64; i++) {
    A[i] = ((s & j) == j);
    j = (j << 1);
  }
  for (i = 0, j = 1; i < 64; i++) {
    B[i] = ((x & j) == j);
    j = (j << 1);
  }
  for (i = 64 - 1; i > -1 && A[i] == 0 && B[i] == 0; i--)
    ;
  if (B[i] == 1 && A[i] == 0) ans = 0;
  for (carry = 0; i > -1 && ans > 0; i--) {
    if (carry == 0) {
      if (A[i] == 1)
        if (B[i] == 1) {
          ans *= 2;
          carry = 0;
        } else {
          carry = 1;
          extra = 0;
        }
      else if (B[i] == 1)
        ans = 0;
      else
        carry = 0;
    } else {
      if (A[i] == 1)
        if (B[i] == 1)
          ans = 0;
        else {
          carry = 1;
          extra = 0;
        }
      else if (B[i] == 1) {
        ans *= 2;
        carry = 1;
      } else {
        carry = 0;
        extra = 0;
      }
    }
  }
  if (carry != 0 || ans == 0) ans = extra = 0;
  cout << ans - extra << '\n';
  return 0;
}

### Prompt

Please provide a CPP coded solution 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 A[64], B[64];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long s, x, i, j, k, ans = 1, carry, extra = 2;
  cin >> s >> x;
  for (i = 0, j = 1; i < 64; i++) {
    A[i] = ((s & j) == j);
    j = (j << 1);
  }
  for (i = 0, j = 1; i < 64; i++) {
    B[i] = ((x & j) == j);
    j = (j << 1);
  }
  for (i = 64 - 1; i > -1 && A[i] == 0 && B[i] == 0; i--)
    ;
  if (B[i] == 1 && A[i] == 0) ans = 0;
  for (carry = 0; i > -1 && ans > 0; i--) {
    if (carry == 0) {
      if (A[i] == 1)
        if (B[i] == 1) {
          ans *= 2;
          carry = 0;
        } else {
          carry = 1;
          extra = 0;
        }
      else if (B[i] == 1)
        ans = 0;
      else
        carry = 0;
    } else {
      if (A[i] == 1)
        if (B[i] == 1)
          ans = 0;
        else {
          carry = 1;
          extra = 0;
        }
      else if (B[i] == 1) {
        ans *= 2;
        carry = 1;
      } else {
        carry = 0;
        extra = 0;
      }
    }
  }
  if (carry != 0 || ans == 0) ans = extra = 0;
  cout << ans - extra << '\n';
  return 0;
}

```