#include <bits/stdc++.h>
using namespace std;
long long s, x;
short check[45];
int main() {
  ios_base::sync_with_stdio(0);
  long long s, x;
  cin >> s >> x;
  int T = 0;
  for (int i = 0; i <= 40; ++i) {
    if (x >> i & 1) {
      check[i] = 1;
      ++T;
      s -= (1ll << i);
    };
  }
  if (s < 0) {
    cout << 0;
    return 0;
  }
  if (s % 2 == 1) {
    cout << 0;
    return 0;
  }
  s >>= 1;
  for (int i = 0; i <= 40; ++i) {
    if (s >> i & 1 && check[i]) {
      cout << 0;
      return 0;
    }
  }
  if (s == x) {
    cout << ((1ll << T) - 2);
    return 0;
  }
  if (s != 0)
    cout << (1ll << T);
  else
    cout << (1ll << T) - 2;
  return 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;
long long s, x;
short check[45];
int main() {
  ios_base::sync_with_stdio(0);
  long long s, x;
  cin >> s >> x;
  int T = 0;
  for (int i = 0; i <= 40; ++i) {
    if (x >> i & 1) {
      check[i] = 1;
      ++T;
      s -= (1ll << i);
    };
  }
  if (s < 0) {
    cout << 0;
    return 0;
  }
  if (s % 2 == 1) {
    cout << 0;
    return 0;
  }
  s >>= 1;
  for (int i = 0; i <= 40; ++i) {
    if (s >> i & 1 && check[i]) {
      cout << 0;
      return 0;
    }
  }
  if (s == x) {
    cout << ((1ll << T) - 2);
    return 0;
  }
  if (s != 0)
    cout << (1ll << T);
  else
    cout << (1ll << T) - 2;
  return 0;
}

```