#include <bits/stdc++.h>
using namespace std;
void init() { return; }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  init();
  long long int s, x;
  cin >> s >> x;
  if (x == 0) {
    if (s % 2) {
      cout << 0 << endl;
    } else
      cout << 1 << endl;
    exit(0);
  }
  int cnt, temp;
  for (int i = 0; i <= 40; i++) {
    if ((s >> i) & 1ll) {
      cnt = i;
    }
    if ((x >> i) & 1ll) {
      temp = i;
    }
  }
  if (temp > cnt) {
    cout << 0 << endl;
    exit(0);
  }
  long long int ans = 1ll;
  int c = 0;
  for (int i = cnt; i >= 0; i--) {
    int a = (s >> i) & 1ll;
    int b = (x >> i) & 1ll;
    if (a == 0 && b == 0) {
      c = 0;
    } else if (a == 1 && b == 0) {
      c = 1;
    } else if (a == 0 && b == 1) {
      if (c == 0) ans = 0;
      c = 1;
      ans = ans * 2ll;
    } else if (a == 1 && b == 1) {
      if (c == 1) ans = 0;
      c = 0;
      ans = ans * 2ll;
    }
  }
  if (c == 1) ans = 0;
  if (s == x) ans -= 2;
  cout << ans << endl;
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
void init() { return; }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  init();
  long long int s, x;
  cin >> s >> x;
  if (x == 0) {
    if (s % 2) {
      cout << 0 << endl;
    } else
      cout << 1 << endl;
    exit(0);
  }
  int cnt, temp;
  for (int i = 0; i <= 40; i++) {
    if ((s >> i) & 1ll) {
      cnt = i;
    }
    if ((x >> i) & 1ll) {
      temp = i;
    }
  }
  if (temp > cnt) {
    cout << 0 << endl;
    exit(0);
  }
  long long int ans = 1ll;
  int c = 0;
  for (int i = cnt; i >= 0; i--) {
    int a = (s >> i) & 1ll;
    int b = (x >> i) & 1ll;
    if (a == 0 && b == 0) {
      c = 0;
    } else if (a == 1 && b == 0) {
      c = 1;
    } else if (a == 0 && b == 1) {
      if (c == 0) ans = 0;
      c = 1;
      ans = ans * 2ll;
    } else if (a == 1 && b == 1) {
      if (c == 1) ans = 0;
      c = 0;
      ans = ans * 2ll;
    }
  }
  if (c == 1) ans = 0;
  if (s == x) ans -= 2;
  cout << ans << endl;
  return 0;
}

```