#include <bits/stdc++.h>
using namespace std;
long long c0[100];
long long c1[100];
void solve() {
  long long s, x;
  long long c = 0;
  cin >> s >> x;
  if (s == x) c = -2;
  long long i = 0;
  if (x % 2 == 1 && s % 2 == 0) {
    cout << 0 << '\n';
    return;
  }
  if (x % 2 == 0 && s % 2 == 1) {
    cout << 0 << '\n';
    return;
  }
  if (s % 2 == 0) {
    c0[1] = 1;
    c1[1] = 1;
  } else {
    c0[1] = 2;
    c1[1] = 0;
  }
  i++;
  x /= 2;
  s /= 2;
  long long r = s % 2;
  while (s != 0 || x != 0) {
    if (x % 2 == 0) {
      if (s % 2 == 0) {
        c0[i + 1] = c0[i];
        c1[i + 1] = c0[i];
      } else {
        c0[i + 1] = c1[i];
        c1[i + 1] = c1[i];
      }
    } else {
      if (s % 2 == 0) {
        c0[i + 1] = 0;
        c1[i + 1] = 2 * c1[i];
      } else {
        c0[i + 1] = 2 * c0[i];
        c1[i + 1] = 0;
      }
    }
    i++;
    r = s % 2;
    s /= 2;
    x /= 2;
  }
  if (x % 2 == 0) {
    if (s % 2 == 0) {
      c0[i + 1] = c0[i];
      c1[i + 1] = c0[i];
    } else {
      c0[i + 1] = c1[i];
      c1[i + 1] = c1[i];
    }
  } else {
    if (s % 2 == 0) {
      c0[i + 1] = 0;
      c1[i + 1] = 2 * c1[i];
    } else {
      c0[i + 1] = 2 * c0[i];
      c1[i + 1] = 0;
    }
  }
  i++;
  r = s % 2;
  s /= 2;
  x /= 2;
  cout << c0[i] + c << '\n';
}
int main() {
  solve();
  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 c0[100];
long long c1[100];
void solve() {
  long long s, x;
  long long c = 0;
  cin >> s >> x;
  if (s == x) c = -2;
  long long i = 0;
  if (x % 2 == 1 && s % 2 == 0) {
    cout << 0 << '\n';
    return;
  }
  if (x % 2 == 0 && s % 2 == 1) {
    cout << 0 << '\n';
    return;
  }
  if (s % 2 == 0) {
    c0[1] = 1;
    c1[1] = 1;
  } else {
    c0[1] = 2;
    c1[1] = 0;
  }
  i++;
  x /= 2;
  s /= 2;
  long long r = s % 2;
  while (s != 0 || x != 0) {
    if (x % 2 == 0) {
      if (s % 2 == 0) {
        c0[i + 1] = c0[i];
        c1[i + 1] = c0[i];
      } else {
        c0[i + 1] = c1[i];
        c1[i + 1] = c1[i];
      }
    } else {
      if (s % 2 == 0) {
        c0[i + 1] = 0;
        c1[i + 1] = 2 * c1[i];
      } else {
        c0[i + 1] = 2 * c0[i];
        c1[i + 1] = 0;
      }
    }
    i++;
    r = s % 2;
    s /= 2;
    x /= 2;
  }
  if (x % 2 == 0) {
    if (s % 2 == 0) {
      c0[i + 1] = c0[i];
      c1[i + 1] = c0[i];
    } else {
      c0[i + 1] = c1[i];
      c1[i + 1] = c1[i];
    }
  } else {
    if (s % 2 == 0) {
      c0[i + 1] = 0;
      c1[i + 1] = 2 * c1[i];
    } else {
      c0[i + 1] = 2 * c0[i];
      c1[i + 1] = 0;
    }
  }
  i++;
  r = s % 2;
  s /= 2;
  x /= 2;
  cout << c0[i] + c << '\n';
}
int main() {
  solve();
  return 0;
}

```