#include <bits/stdc++.h>
using namespace std;
long long int i, j, mini = 1000000000, p, d, n, m, k, rasp, ok, a, b, ul, c, op,
                    maxi;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> m;
  rasp = 1;
  for (i = 0; i < 60; i++) {
    if (m & ((long long int)1 << i)) rasp *= 2;
  }
  n -= m;
  if (n < 0 || n & 1) {
    cout << 0;
    return 0;
  }
  if (n == 0) rasp -= 2;
  for (i = 0; i < 60; i++) {
    if ((1ll << (i + 1)) & n)
      if ((1ll << i) & m) {
        cout << 0;
        return 0;
      }
  }
  cout << rasp;
  return 0;
}

### Prompt

Your task is to create 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;
long long int i, j, mini = 1000000000, p, d, n, m, k, rasp, ok, a, b, ul, c, op,
                    maxi;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> m;
  rasp = 1;
  for (i = 0; i < 60; i++) {
    if (m & ((long long int)1 << i)) rasp *= 2;
  }
  n -= m;
  if (n < 0 || n & 1) {
    cout << 0;
    return 0;
  }
  if (n == 0) rasp -= 2;
  for (i = 0; i < 60; i++) {
    if ((1ll << (i + 1)) & n)
      if ((1ll << i) & m) {
        cout << 0;
        return 0;
      }
  }
  cout << rasp;
  return 0;
}

```