#include <bits/stdc++.h>
using namespace std;
long long x, s;
vector<int> dx, ds;
bool done[51][2];
long long f[51][2];
long long F(int u, int b) {
  if (done[u][b]) return f[u][b];
  done[u][b] = 1;
  if (u == 50) return f[u][b] = (b == 0);
  for (int x = 0; x < 2; x++) {
    int y = dx[u] ^ x;
    if ((x + y + b) % 2 == ds[u]) f[u][b] += F(u + 1, (x + y + b) / 2);
  }
  return f[u][b];
}
int main() {
  cin >> s >> x;
  int minus = (s == x) ? 2 : 0;
  for (int i = 0; i < 50; i++) {
    dx.push_back(x % 2);
    x /= 2;
  }
  for (int i = 0; i < 50; i++) {
    ds.push_back(s % 2);
    s /= 2;
  }
  cout << F(0, 0) - minus << '\n';
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 x, s;
vector<int> dx, ds;
bool done[51][2];
long long f[51][2];
long long F(int u, int b) {
  if (done[u][b]) return f[u][b];
  done[u][b] = 1;
  if (u == 50) return f[u][b] = (b == 0);
  for (int x = 0; x < 2; x++) {
    int y = dx[u] ^ x;
    if ((x + y + b) % 2 == ds[u]) f[u][b] += F(u + 1, (x + y + b) / 2);
  }
  return f[u][b];
}
int main() {
  cin >> s >> x;
  int minus = (s == x) ? 2 : 0;
  for (int i = 0; i < 50; i++) {
    dx.push_back(x % 2);
    x /= 2;
  }
  for (int i = 0; i < 50; i++) {
    ds.push_back(s % 2);
    s /= 2;
  }
  cout << F(0, 0) - minus << '\n';
}

```