#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long tot, xo;
  cin >> tot >> xo;
  long long pw = 1;
  bool flag[50];
  memset(flag, 0, sizeof flag);
  int cnt = 0;
  for (int i = 0; i < 45; i++) {
    if (pw & xo) {
      tot -= pw;
      flag[i] = true;
      cnt++;
    }
    pw *= 2;
  }
  bool valid = true;
  if (tot & 1) valid = false;
  tot /= 2;
  pw = 1;
  for (int i = 0; i < 45; i++) {
    if (flag[i] && (tot & pw)) valid = false;
    pw *= 2;
  }
  long long ans = 0;
  if (valid) {
    ans = (long long)1 << cnt;
    if (tot == 0) ans -= 2;
  }
  cout << ans << endl;
  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;
int main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long tot, xo;
  cin >> tot >> xo;
  long long pw = 1;
  bool flag[50];
  memset(flag, 0, sizeof flag);
  int cnt = 0;
  for (int i = 0; i < 45; i++) {
    if (pw & xo) {
      tot -= pw;
      flag[i] = true;
      cnt++;
    }
    pw *= 2;
  }
  bool valid = true;
  if (tot & 1) valid = false;
  tot /= 2;
  pw = 1;
  for (int i = 0; i < 45; i++) {
    if (flag[i] && (tot & pw)) valid = false;
    pw *= 2;
  }
  long long ans = 0;
  if (valid) {
    ans = (long long)1 << cnt;
    if (tot == 0) ans -= 2;
  }
  cout << ans << endl;
  return 0;
}

```