#include <bits/stdc++.h>
using namespace std;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long n, m;
  cin >> n >> m;
  long long mm = 0;
  if (n == m) mm = 1;
  long long flag = 0;
  if ((n - m) & 1) flag = 1;
  n = (n - m) / 2;
  long long res = 1;
  for (long long i = 0; i < 60; i++) {
    if ((1LL << i) & n) {
      if ((1LL << i) & m) flag = 1;
    } else {
      if ((1LL << i) & m) res *= 2;
    }
  }
  res -= 2 * mm;
  if (flag)
    cout << "0\n";
  else
    cout << res << "\n";
  return 0;
}

### Prompt

Develop a solution in Cpp 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;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long n, m;
  cin >> n >> m;
  long long mm = 0;
  if (n == m) mm = 1;
  long long flag = 0;
  if ((n - m) & 1) flag = 1;
  n = (n - m) / 2;
  long long res = 1;
  for (long long i = 0; i < 60; i++) {
    if ((1LL << i) & n) {
      if ((1LL << i) & m) flag = 1;
    } else {
      if ((1LL << i) & m) res *= 2;
    }
  }
  res -= 2 * mm;
  if (flag)
    cout << "0\n";
  else
    cout << res << "\n";
  return 0;
}

```