#include <bits/stdc++.h>
using namespace std;
long long dp[2], dp2[2];
int main() {
  long long x, y;
  scanf("%I64d%I64d", &x, &y);
  int ans = 0;
  if (x == y) ans = -2;
  dp[0] = 1;
  dp[1] = 0;
  for (int i = 0; i < 40; i++) {
    int p = x & 1, q = y & 1;
    x >>= 1;
    y >>= 1;
    dp2[0] = dp2[1] = 0;
    for (int j = 0; j < 2; j++)
      for (int k = 0; k < 2; k++)
        for (int l = 0; l < 2; l++)
          if ((k ^ l == q) && (k ^ l ^ j == p)) dp2[(k + l + j) >> 1] += dp[j];
    dp[0] = dp2[0];
    dp[1] = dp2[1];
  }
  printf("%I64d\n", dp[0] + ans);
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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 dp[2], dp2[2];
int main() {
  long long x, y;
  scanf("%I64d%I64d", &x, &y);
  int ans = 0;
  if (x == y) ans = -2;
  dp[0] = 1;
  dp[1] = 0;
  for (int i = 0; i < 40; i++) {
    int p = x & 1, q = y & 1;
    x >>= 1;
    y >>= 1;
    dp2[0] = dp2[1] = 0;
    for (int j = 0; j < 2; j++)
      for (int k = 0; k < 2; k++)
        for (int l = 0; l < 2; l++)
          if ((k ^ l == q) && (k ^ l ^ j == p)) dp2[(k + l + j) >> 1] += dp[j];
    dp[0] = dp2[0];
    dp[1] = dp2[1];
  }
  printf("%I64d\n", dp[0] + ans);
  return 0;
}

```