#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 10, inf = 1e9 + 10, P = 53, MOD = 1e9 + 7;
vector<int> a, b;
int main() {
  ios_base::sync_with_stdio(false);
  long long n, x, t = 0;
  long long ans = 1;
  cin >> n >> x;
  for (int i = 0; i < 42; i++)
    if ((1ll << i) & x) {
      ans *= 2;
      n -= (1ll << i);
      t++;
    }
  if (n < 0 || (n == 0 && t == 1)) {
    cout << 0;
    return 0;
  }
  if (n == 0) {
    cout << ans - 2;
    return 0;
  }
  for (int i = 0; i < 42; i++)
    if (!((1ll << i) & x))
      if ((1ll << (i + 1)) & n) n -= (1ll << (i + 1));
  if (n == 0)
    cout << ans;
  else
    cout << 0;
  return 0;
}

### 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;
const int maxn = 2e5 + 10, inf = 1e9 + 10, P = 53, MOD = 1e9 + 7;
vector<int> a, b;
int main() {
  ios_base::sync_with_stdio(false);
  long long n, x, t = 0;
  long long ans = 1;
  cin >> n >> x;
  for (int i = 0; i < 42; i++)
    if ((1ll << i) & x) {
      ans *= 2;
      n -= (1ll << i);
      t++;
    }
  if (n < 0 || (n == 0 && t == 1)) {
    cout << 0;
    return 0;
  }
  if (n == 0) {
    cout << ans - 2;
    return 0;
  }
  for (int i = 0; i < 42; i++)
    if (!((1ll << i) & x))
      if ((1ll << (i + 1)) & n) n -= (1ll << (i + 1));
  if (n == 0)
    cout << ans;
  else
    cout << 0;
  return 0;
}

```