#include <bits/stdc++.h>
using namespace std;
long long int power(int x) {
  long long int t = 1;
  while (x--) t = t * 2;
  return t;
}
int main() {
  long long int x, s;
  cin >> s >> x;
  long long int temp = x, ans = 0, flag = 0;
  if (x == 0) {
    if (s % 2 == 0)
      cout << 1 << endl;
    else
      cout << 0 << endl;
    return 0;
  }
  s = s - x;
  if (s % 2 != 0) {
    cout << 0 << endl;
    return 0;
  }
  if (s == 0) flag = 1;
  s = s / 2;
  while (s != 0) {
    long long int t1, t2;
    t1 = s & 1;
    t2 = temp & 1;
    if (t1 == 1 && t2 == 1) {
      break;
    }
    s = s >> 1;
    temp = temp >> 1;
  }
  if (s == 0) {
    long long int count = 0;
    while (x != 0) {
      if (x & 1 == 1) count++;
      x = x >> 1;
    }
    ans = power(count);
    if (flag == 1)
      cout << ans - 2 << endl;
    else
      cout << ans << endl;
  } else
    cout << 0 << endl;
  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;
long long int power(int x) {
  long long int t = 1;
  while (x--) t = t * 2;
  return t;
}
int main() {
  long long int x, s;
  cin >> s >> x;
  long long int temp = x, ans = 0, flag = 0;
  if (x == 0) {
    if (s % 2 == 0)
      cout << 1 << endl;
    else
      cout << 0 << endl;
    return 0;
  }
  s = s - x;
  if (s % 2 != 0) {
    cout << 0 << endl;
    return 0;
  }
  if (s == 0) flag = 1;
  s = s / 2;
  while (s != 0) {
    long long int t1, t2;
    t1 = s & 1;
    t2 = temp & 1;
    if (t1 == 1 && t2 == 1) {
      break;
    }
    s = s >> 1;
    temp = temp >> 1;
  }
  if (s == 0) {
    long long int count = 0;
    while (x != 0) {
      if (x & 1 == 1) count++;
      x = x >> 1;
    }
    ans = power(count);
    if (flag == 1)
      cout << ans - 2 << endl;
    else
      cout << ans << endl;
  } else
    cout << 0 << endl;
  return 0;
}

```