#include <bits/stdc++.h>
using namespace std;
int A[51], B[51], C[51];
int maxb;
void get(long long p, int X[]) {
  int cnt = 50;
  maxb = 0;
  while (p) {
    X[cnt--] = p % 2;
    p /= 2;
    maxb++;
  }
  maxb = 50 - maxb;
}
int main() {
  long long s, x;
  cin >> s >> x;
  if (x > s) {
    printf("0");
    return 0;
  }
  get(s, A);
  int lim = maxb;
  maxb = 0;
  get(x, B);
  int i;
  for (i = 50; i >= lim; i--) {
    C[i] = A[i] ^ B[i];
  }
  if (C[50] == 1) {
    cout << "0";
    return 0;
  }
  long long ans = 1;
  for (i = 50; i >= lim; i--) {
    if (C[i - 1] == 1) {
      if (B[i] == 0) {
        continue;
      } else {
        if (C[i] == 0)
          ans = 0;
        else
          ans *= 2;
      }
    } else {
      if (B[i] == 1) {
        if (C[i])
          ans = 0;
        else
          ans *= 2;
      }
    }
  }
  if (s == x) ans -= 2;
  cout << ans;
}

### 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 A[51], B[51], C[51];
int maxb;
void get(long long p, int X[]) {
  int cnt = 50;
  maxb = 0;
  while (p) {
    X[cnt--] = p % 2;
    p /= 2;
    maxb++;
  }
  maxb = 50 - maxb;
}
int main() {
  long long s, x;
  cin >> s >> x;
  if (x > s) {
    printf("0");
    return 0;
  }
  get(s, A);
  int lim = maxb;
  maxb = 0;
  get(x, B);
  int i;
  for (i = 50; i >= lim; i--) {
    C[i] = A[i] ^ B[i];
  }
  if (C[50] == 1) {
    cout << "0";
    return 0;
  }
  long long ans = 1;
  for (i = 50; i >= lim; i--) {
    if (C[i - 1] == 1) {
      if (B[i] == 0) {
        continue;
      } else {
        if (C[i] == 0)
          ans = 0;
        else
          ans *= 2;
      }
    } else {
      if (B[i] == 1) {
        if (C[i])
          ans = 0;
        else
          ans *= 2;
      }
    }
  }
  if (s == x) ans -= 2;
  cout << ans;
}

```