#include <bits/stdc++.h>
using namespace std;
long long dp[200][2];
int main() {
  long long s, x;
  scanf("%I64d %I64d", &s, &x);
  if (s < x) {
    printf("0\n");
    return 0;
  }
  int i = 0, ls, lx;
  long long cs = s, cx = x;
  dp[0][0] = 1;
  while (s > 0LL) {
    ls = s % 2;
    s /= 2;
    lx = x % 2;
    x /= 2;
    if (ls == 0 && lx == 1) {
      dp[i + 1][1] += 2 * dp[i][1];
    }
    if (ls == 0 && lx == 0) {
      dp[i + 1][1] += dp[i][0];
      dp[i + 1][0] += dp[i][0];
    }
    if (ls == 1 && lx == 1) {
      dp[i + 1][0] += 2 * dp[i][0];
    }
    if (ls == 1 && lx == 0) {
      dp[i + 1][1] += dp[i][1];
      dp[i + 1][0] += dp[i][1];
    }
    i++;
  }
  if (cs == cx) dp[i][0] -= 2;
  printf("%I64d\n", dp[i][0]);
  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[200][2];
int main() {
  long long s, x;
  scanf("%I64d %I64d", &s, &x);
  if (s < x) {
    printf("0\n");
    return 0;
  }
  int i = 0, ls, lx;
  long long cs = s, cx = x;
  dp[0][0] = 1;
  while (s > 0LL) {
    ls = s % 2;
    s /= 2;
    lx = x % 2;
    x /= 2;
    if (ls == 0 && lx == 1) {
      dp[i + 1][1] += 2 * dp[i][1];
    }
    if (ls == 0 && lx == 0) {
      dp[i + 1][1] += dp[i][0];
      dp[i + 1][0] += dp[i][0];
    }
    if (ls == 1 && lx == 1) {
      dp[i + 1][0] += 2 * dp[i][0];
    }
    if (ls == 1 && lx == 0) {
      dp[i + 1][1] += dp[i][1];
      dp[i + 1][0] += dp[i][1];
    }
    i++;
  }
  if (cs == cx) dp[i][0] -= 2;
  printf("%I64d\n", dp[i][0]);
  return 0;
}

```