#include <bits/stdc++.h>
using namespace std;
long long p(long long x, long long y);
int main() {
  long long s, x;
  cin >> s >> x;
  long long y = (((s - x) & 1) == 0 ? (s - x) >> 1 : -1);
  if (y == -1) {
    cout << 0;
    return 0;
  }
  for (int i = 40; i >= 0; i--) {
    if (((1LL << i) & x) && ((1LL << i) & y)) {
      cout << 0;
      return 0;
    }
  }
  int idx1 = (int)(log2(x * 1.0));
  long long bits = 0;
  for (int i = idx1; i >= 0; i--) {
    if ((1LL << i) & x) bits++;
  }
  long long ans = p(2LL, bits);
  if (y == 0) ans -= 2;
  cout << ans;
  return 0;
}
long long p(long long x, long long y) {
  if (y == 0) return 1LL;
  if (y == 1) return x;
  long long half = p(x, y / 2);
  if (y % 2 == 0) {
    return half * half;
  } else {
    return half * half * x;
  }
}

### 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 p(long long x, long long y);
int main() {
  long long s, x;
  cin >> s >> x;
  long long y = (((s - x) & 1) == 0 ? (s - x) >> 1 : -1);
  if (y == -1) {
    cout << 0;
    return 0;
  }
  for (int i = 40; i >= 0; i--) {
    if (((1LL << i) & x) && ((1LL << i) & y)) {
      cout << 0;
      return 0;
    }
  }
  int idx1 = (int)(log2(x * 1.0));
  long long bits = 0;
  for (int i = idx1; i >= 0; i--) {
    if ((1LL << i) & x) bits++;
  }
  long long ans = p(2LL, bits);
  if (y == 0) ans -= 2;
  cout << ans;
  return 0;
}
long long p(long long x, long long y) {
  if (y == 0) return 1LL;
  if (y == 1) return x;
  long long half = p(x, y / 2);
  if (y % 2 == 0) {
    return half * half;
  } else {
    return half * half * x;
  }
}

```