#include <bits/stdc++.h>
using namespace std;
const int N = 48;
long long solve(const bitset<N> x, const bitset<N> d) {
  for (int i = N - 1; i > 0; i--)
    if (d[i] && x[i - 1]) return 0;
  return d[0] ? 0 : 1;
}
int main() {
  long long s, x;
  scanf("%I64d%I64d", &s, &x);
  long long ans = 1LL << bitset<N>(x).count();
  if (s < x)
    ans = 0;
  else
    ans *= solve(x, s - x);
  if (ans > 0 && s == x) ans -= 2;
  printf("%I64d\n", ans);
}

### 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;
const int N = 48;
long long solve(const bitset<N> x, const bitset<N> d) {
  for (int i = N - 1; i > 0; i--)
    if (d[i] && x[i - 1]) return 0;
  return d[0] ? 0 : 1;
}
int main() {
  long long s, x;
  scanf("%I64d%I64d", &s, &x);
  long long ans = 1LL << bitset<N>(x).count();
  if (s < x)
    ans = 0;
  else
    ans *= solve(x, s - x);
  if (ans > 0 && s == x) ans -= 2;
  printf("%I64d\n", ans);
}

```