#include <bits/stdc++.h>
using namespace std;
long long s, x;
int main() {
  scanf("%lld%lld", &s, &x);
  int sub = 0;
  if (s == x) sub = 2;
  int pop = __builtin_popcountll(x);
  for (int i = 0; i < 60; i++)
    if ((x >> i) & 1) s -= 1LL << i;
  for (int i = 59; i >= 0; i--)
    if (!((x >> i) & 1) && s >= (1LL << (i + 1))) s -= 1LL << (i + 1);
  long long ans = 1LL << pop;
  if (s == 0)
    printf("%lld\n", ans - sub);
  else
    printf("0\n");
  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;
long long s, x;
int main() {
  scanf("%lld%lld", &s, &x);
  int sub = 0;
  if (s == x) sub = 2;
  int pop = __builtin_popcountll(x);
  for (int i = 0; i < 60; i++)
    if ((x >> i) & 1) s -= 1LL << i;
  for (int i = 59; i >= 0; i--)
    if (!((x >> i) & 1) && s >= (1LL << (i + 1))) s -= 1LL << (i + 1);
  long long ans = 1LL << pop;
  if (s == 0)
    printf("%lld\n", ans - sub);
  else
    printf("0\n");
  return 0;
}

```