#include <bits/stdc++.h>
using namespace std;
int i, j, k, n, m;
int a[100];
long long x, s, an;
inline void fail() {
  printf("0\n");
  exit(0);
}
int main() {
  scanf("%I64d%I64d", &x, &s);
  x -= s;
  an = 1;
  if (x < 0) fail();
  for (i = 0; i <= 60; i++)
    if ((s >> i) & 1)
      a[i + 1] = 0, an *= 2;
    else
      a[i + 1] = 1;
  for (i = 0; i <= 60; i++)
    if ((x >> i) & 1) {
      if (!a[i]) fail();
    }
  if (x == 0) an -= 2;
  printf("%I64d\n", an);
  return 0;
}

### Prompt

Your task is to create 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;
int i, j, k, n, m;
int a[100];
long long x, s, an;
inline void fail() {
  printf("0\n");
  exit(0);
}
int main() {
  scanf("%I64d%I64d", &x, &s);
  x -= s;
  an = 1;
  if (x < 0) fail();
  for (i = 0; i <= 60; i++)
    if ((s >> i) & 1)
      a[i + 1] = 0, an *= 2;
    else
      a[i + 1] = 1;
  for (i = 0; i <= 60; i++)
    if ((x >> i) & 1) {
      if (!a[i]) fail();
    }
  if (x == 0) an -= 2;
  printf("%I64d\n", an);
  return 0;
}

```