#include <bits/stdc++.h>
using namespace std;
long long s, x, a[100], pw[45];
void calc() {
  pw[0] = 1;
  for (long long i = 1; i <= 40; i++) pw[i] = (pw[i - 1] * 2);
}
int main() {
  calc();
  long long i, j, n, m, temp = 0, rem, f_ans = 1, cnt = 0;
  bool ans = true;
  cin >> s >> x;
  for (i = 0; i <= 40; i++) {
    m = x & pw[i];
    if (m > 0)
      a[i] = 1, temp += pw[i];
    else
      a[i] = 0;
  }
  rem = s - temp;
  if (rem % 2 == 1) ans = false;
  rem /= 2;
  for (i = 0; i <= 40; i++) {
    m = rem & pw[i];
    if (m > 0 && a[i] == 1) ans = false;
  }
  if (ans) {
    for (i = 0; i <= 40; i++)
      if (a[i] == 1) f_ans *= 2;
  }
  if (ans && rem > 0)
    cout << f_ans;
  else if (ans && rem == 0)
    cout << f_ans - 2;
  else
    cout << "0";
  return 0;
}

### Prompt

Your challenge is to write a Cpp solution 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;
long long s, x, a[100], pw[45];
void calc() {
  pw[0] = 1;
  for (long long i = 1; i <= 40; i++) pw[i] = (pw[i - 1] * 2);
}
int main() {
  calc();
  long long i, j, n, m, temp = 0, rem, f_ans = 1, cnt = 0;
  bool ans = true;
  cin >> s >> x;
  for (i = 0; i <= 40; i++) {
    m = x & pw[i];
    if (m > 0)
      a[i] = 1, temp += pw[i];
    else
      a[i] = 0;
  }
  rem = s - temp;
  if (rem % 2 == 1) ans = false;
  rem /= 2;
  for (i = 0; i <= 40; i++) {
    m = rem & pw[i];
    if (m > 0 && a[i] == 1) ans = false;
  }
  if (ans) {
    for (i = 0; i <= 40; i++)
      if (a[i] == 1) f_ans *= 2;
  }
  if (ans && rem > 0)
    cout << f_ans;
  else if (ans && rem == 0)
    cout << f_ans - 2;
  else
    cout << "0";
  return 0;
}

```