#include <bits/stdc++.h>
using namespace std;
const long long MAXN = 1e13;
bool v[40];
int main() {
  long long sum, xr, ways = 1, origsum, origxr;
  cin >> sum >> xr;
  origsum = sum, origxr = xr;
  for (long long i = 1, j = 0; i < MAXN; i *= 2, j++) {
    if (xr & i) {
      sum -= i;
      ways *= 2;
    } else
      v[j] = true;
  }
  if (origsum == origxr) ways -= 2;
  for (long long i = 2, j = 0; i < MAXN; i *= 2, j++) {
    if (sum & i) {
      if (!v[j])
        ways = 0;
      else
        sum -= i;
    }
  }
  if (sum > 0) ways = 0;
  cout << ways << endl;
}

### 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;
const long long MAXN = 1e13;
bool v[40];
int main() {
  long long sum, xr, ways = 1, origsum, origxr;
  cin >> sum >> xr;
  origsum = sum, origxr = xr;
  for (long long i = 1, j = 0; i < MAXN; i *= 2, j++) {
    if (xr & i) {
      sum -= i;
      ways *= 2;
    } else
      v[j] = true;
  }
  if (origsum == origxr) ways -= 2;
  for (long long i = 2, j = 0; i < MAXN; i *= 2, j++) {
    if (sum & i) {
      if (!v[j])
        ways = 0;
      else
        sum -= i;
    }
  }
  if (sum > 0) ways = 0;
  cout << ways << endl;
}

```