#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
using ii = pair<ll, ll>;
using vi = vector<ll>;
using vb = vector<bool>;
using vvi = vector<vi>;
using vii = vector<ii>;
using vvii = vector<vii>;
const int INF = 2000000000;
const ll LLINF = 9000000000000000000;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  ll S, X;
  cin >> S >> X;
  S -= X;
  if ((S & 1) || S < 0) {
    cout << 0 << endl;
    return 0;
  }
  cerr << S << ' ' << X << endl;
  S /= 2LL;
  int exp = __builtin_popcountll(X & ~S);
  ll ans = 1LL << exp;
  if (S == 0) ans -= 2LL;
  if (ans < 0LL || ((X & S) != 0LL)) ans *= 0LL;
  cout << ans << endl;
  return 0;
}

### Prompt

Generate 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;
using ll = long long;
using ld = long double;
using ii = pair<ll, ll>;
using vi = vector<ll>;
using vb = vector<bool>;
using vvi = vector<vi>;
using vii = vector<ii>;
using vvii = vector<vii>;
const int INF = 2000000000;
const ll LLINF = 9000000000000000000;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  ll S, X;
  cin >> S >> X;
  S -= X;
  if ((S & 1) || S < 0) {
    cout << 0 << endl;
    return 0;
  }
  cerr << S << ' ' << X << endl;
  S /= 2LL;
  int exp = __builtin_popcountll(X & ~S);
  ll ans = 1LL << exp;
  if (S == 0) ans -= 2LL;
  if (ans < 0LL || ((X & S) != 0LL)) ans *= 0LL;
  cout << ans << endl;
  return 0;
}

```