#include <bits/stdc++.h>
using namespace std;
int main() {
  long long int nn, mm, cnt = 0, c = 0, ans = 1, i, j, k, l, n, m, a[100] = {0},
                        b[100] = {0};
  cin >> n >> m;
  i = j = 0;
  nn = n;
  mm = m;
  while (n) {
    a[i++] = n % 2;
    n /= 2;
  }
  while (m) {
    b[j++] = m % 2;
    m /= 2;
  }
  for (k = 0; k < j; k++)
    if (b[k]) {
      ans *= 2;
      cnt++;
    }
  for (k = 0; k < j || k < i; k++)
    if (b[k]) {
      if (c) {
        if (a[k] == 1) {
          cout << "0\n";
          return 0;
        }
      } else if (!a[k]) {
        cout << "0\n";
        return 0;
      }
    } else {
      if (c) {
        if (!a[k]) {
          cout << "0\n";
          return 0;
        }
      } else if (a[k]) {
        cout << "0\n";
        return 0;
      }
      if (b[k + 1])
        if (a[k + 1])
          c = 0;
        else
          c = 1;
      else if (a[k + 1])
        c = 1;
      else
        c = 0;
    }
  if (c)
    cout << "0\n";
  else {
    if (nn == mm) ans -= 2;
    cout << ans << endl;
  }
  return 0;
}

### Prompt

In Cpp, your task is to solve 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 main() {
  long long int nn, mm, cnt = 0, c = 0, ans = 1, i, j, k, l, n, m, a[100] = {0},
                        b[100] = {0};
  cin >> n >> m;
  i = j = 0;
  nn = n;
  mm = m;
  while (n) {
    a[i++] = n % 2;
    n /= 2;
  }
  while (m) {
    b[j++] = m % 2;
    m /= 2;
  }
  for (k = 0; k < j; k++)
    if (b[k]) {
      ans *= 2;
      cnt++;
    }
  for (k = 0; k < j || k < i; k++)
    if (b[k]) {
      if (c) {
        if (a[k] == 1) {
          cout << "0\n";
          return 0;
        }
      } else if (!a[k]) {
        cout << "0\n";
        return 0;
      }
    } else {
      if (c) {
        if (!a[k]) {
          cout << "0\n";
          return 0;
        }
      } else if (a[k]) {
        cout << "0\n";
        return 0;
      }
      if (b[k + 1])
        if (a[k + 1])
          c = 0;
        else
          c = 1;
      else if (a[k + 1])
        c = 1;
      else
        c = 0;
    }
  if (c)
    cout << "0\n";
  else {
    if (nn == mm) ans -= 2;
    cout << ans << endl;
  }
  return 0;
}

```