#include <bits/stdc++.h>
using namespace std;
template <class A, class B>
inline bool mina(A &x, const B &y) {
  return (y < x) ? (x = y, 1) : 0;
}
template <class A, class B>
inline bool maxa(A &x, const B &y) {
  return (x < y) ? (x = y, 1) : 0;
}
int main() {
  {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
  };
  long long S, X;
  cin >> S >> X;
  long long res;
  if (S < X)
    res = 0;
  else if (S == X)
    res = (1LL << __builtin_popcountll(X)) - 2;
  else {
    int c = 0;
    res = 1;
    for (int _a = (-1), i = (50); i > _a; --i) {
      int a = (S >> i) & 1;
      int b = (X >> i) & 1;
      if (b == 1) {
        if (a == c)
          res = 0;
        else
          res <<= 1;
      } else
        c = a;
    }
    if (c == 1) res = 0;
  }
  cout << res << '\n';
  return 0;
}

### Prompt

Please formulate 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;
template <class A, class B>
inline bool mina(A &x, const B &y) {
  return (y < x) ? (x = y, 1) : 0;
}
template <class A, class B>
inline bool maxa(A &x, const B &y) {
  return (x < y) ? (x = y, 1) : 0;
}
int main() {
  {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
  };
  long long S, X;
  cin >> S >> X;
  long long res;
  if (S < X)
    res = 0;
  else if (S == X)
    res = (1LL << __builtin_popcountll(X)) - 2;
  else {
    int c = 0;
    res = 1;
    for (int _a = (-1), i = (50); i > _a; --i) {
      int a = (S >> i) & 1;
      int b = (X >> i) & 1;
      if (b == 1) {
        if (a == c)
          res = 0;
        else
          res <<= 1;
      } else
        c = a;
    }
    if (c == 1) res = 0;
  }
  cout << res << '\n';
  return 0;
}

```