#include <bits/stdc++.h>
using namespace std;
long long int dp[70][2];
int main() {
  long long S, X;
  scanf("%I64d%I64d", &S, &X);
  long long s = S, x = X;
  dp[0][0] = 1ll;
  dp[0][1] = 0ll;
  int i = 1;
  long long sb, xb;
  while (s || x) {
    sb = s & 1ll, xb = x & 1ll;
    if ((!sb) && (!xb)) {
      dp[i][0] = dp[i - 1][0];
      dp[i][1] = dp[i - 1][0];
    } else if ((!sb) && xb) {
      dp[i][0] = 0;
      dp[i][1] = 2 * dp[i - 1][1];
    } else if (sb && xb) {
      dp[i][0] = 2 * dp[i - 1][0];
      dp[i][1] = 0;
    } else {
      dp[i][0] = dp[i - 1][1];
      dp[i][1] = dp[i - 1][1];
    }
    i++;
    s >>= 1;
    x >>= 1;
  }
  if ((S == X) && (S)) dp[i - 1][0] -= 2ll;
  printf("%I64d\n", dp[i - 1][0]);
  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 int dp[70][2];
int main() {
  long long S, X;
  scanf("%I64d%I64d", &S, &X);
  long long s = S, x = X;
  dp[0][0] = 1ll;
  dp[0][1] = 0ll;
  int i = 1;
  long long sb, xb;
  while (s || x) {
    sb = s & 1ll, xb = x & 1ll;
    if ((!sb) && (!xb)) {
      dp[i][0] = dp[i - 1][0];
      dp[i][1] = dp[i - 1][0];
    } else if ((!sb) && xb) {
      dp[i][0] = 0;
      dp[i][1] = 2 * dp[i - 1][1];
    } else if (sb && xb) {
      dp[i][0] = 2 * dp[i - 1][0];
      dp[i][1] = 0;
    } else {
      dp[i][0] = dp[i - 1][1];
      dp[i][1] = dp[i - 1][1];
    }
    i++;
    s >>= 1;
    x >>= 1;
  }
  if ((S == X) && (S)) dp[i - 1][0] -= 2ll;
  printf("%I64d\n", dp[i - 1][0]);
  return 0;
}

```