#include <bits/stdc++.h>
using namespace std;
inline long long in() {
  int32_t x;
  scanf("%d", &x);
  return x;
}
inline long long lin() {
  long long x;
  scanf("%lld", &x);
  return x;
}
const long long maxn = 1e6 + 10;
int32_t main() {
  long long sum, xo;
  cin >> sum >> xo;
  long long ways = 1;
  bool fl = sum == xo;
  for (long long bit = 41; bit >= 0; bit--) {
    long long val = 1LL << bit;
    if (xo & val) {
      xo -= val;
      ways *= 2;
      sum -= val;
      continue;
    }
    if (sum >= 2 * val) {
      sum -= 2 * val;
    }
  }
  if (sum == 0 && xo == 0) return cout << ways - 2 * fl << endl, 0;
  cout << 0 << endl;
}

### Prompt

Please formulate 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;
inline long long in() {
  int32_t x;
  scanf("%d", &x);
  return x;
}
inline long long lin() {
  long long x;
  scanf("%lld", &x);
  return x;
}
const long long maxn = 1e6 + 10;
int32_t main() {
  long long sum, xo;
  cin >> sum >> xo;
  long long ways = 1;
  bool fl = sum == xo;
  for (long long bit = 41; bit >= 0; bit--) {
    long long val = 1LL << bit;
    if (xo & val) {
      xo -= val;
      ways *= 2;
      sum -= val;
      continue;
    }
    if (sum >= 2 * val) {
      sum -= 2 * val;
    }
  }
  if (sum == 0 && xo == 0) return cout << ways - 2 * fl << endl, 0;
  cout << 0 << endl;
}

```