#include <bits/stdc++.h>
using namespace std;
void FFF() {
  cout.precision(18);
  cout << fixed;
}
void FST() {
  cin.tie(0);
  cout.tie(0);
  ios_base::sync_with_stdio(0);
}
long long n, m, i, j, kol;
int main() {
  cin >> n >> m;
  if (n - m < 0 || (n - m) % 2 == 1 || (m & ((n - m) / 2)) != 0) {
    cout << 0;
    return 0;
  }
  j = (n - m) / 2;
  while (m > 0) {
    if (m % 2 == 1) ++kol;
    m >>= 1;
  }
  if (j == 0)
    cout << (1LL << kol) - 2;
  else
    cout << (1LL << kol);
}

### Prompt

Generate 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;
void FFF() {
  cout.precision(18);
  cout << fixed;
}
void FST() {
  cin.tie(0);
  cout.tie(0);
  ios_base::sync_with_stdio(0);
}
long long n, m, i, j, kol;
int main() {
  cin >> n >> m;
  if (n - m < 0 || (n - m) % 2 == 1 || (m & ((n - m) / 2)) != 0) {
    cout << 0;
    return 0;
  }
  j = (n - m) / 2;
  while (m > 0) {
    if (m % 2 == 1) ++kol;
    m >>= 1;
  }
  if (j == 0)
    cout << (1LL << kol) - 2;
  else
    cout << (1LL << kol);
}

```