#include <bits/stdc++.h>
using namespace std;
int main() {
  long long a, b;
  cin >> a >> b;
  long long two_ones = (a - b) >> 1LL;
  long long shift = 0;
  long long num1 = 0;
  long long num2 = 0;
  for (long long i = 0; i < 64; i++) {
    if ((b >> i) & 1LL) {
      num1 |= (1LL << i);
      shift++;
    } else if ((two_ones >> i) & 1LL) {
      num1 |= (1LL << i);
      num2 |= (1LL << i);
    }
  }
  if (num1 + num2 == a && !(two_ones & b)) {
    long long ans = 1LL << shift;
    if (num2 == 0) ans -= 2;
    cout << ans;
  } else
    cout << 0;
}

### Prompt

In Cpp, your task is to solve 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;
int main() {
  long long a, b;
  cin >> a >> b;
  long long two_ones = (a - b) >> 1LL;
  long long shift = 0;
  long long num1 = 0;
  long long num2 = 0;
  for (long long i = 0; i < 64; i++) {
    if ((b >> i) & 1LL) {
      num1 |= (1LL << i);
      shift++;
    } else if ((two_ones >> i) & 1LL) {
      num1 |= (1LL << i);
      num2 |= (1LL << i);
    }
  }
  if (num1 + num2 == a && !(two_ones & b)) {
    long long ans = 1LL << shift;
    if (num2 == 0) ans -= 2;
    cout << ans;
  } else
    cout << 0;
}

```