#include <bits/stdc++.h>
using namespace std;
long long s, x;
long long solve() {
  long long v = 0;
  for (int i = 0; i < 60; i++) {
    if ((x & (1LL << i)) != 0) {
      v++;
    }
  }
  if (s < x) {
    return 0LL;
  } else if (s == x) {
    return (1LL << v) - 2;
  } else {
    long long d = s - x;
    if ((d & 1) == 1) {
      return 0LL;
    } else {
      for (int i = 0; i < 60; i++) {
        if ((d & (1LL << i)) != 0 && (x & (1LL << (i - 1))) != 0) {
          return 0LL;
        }
      }
      return (1LL << v);
    }
  }
}
int main() {
  scanf("%I64d %I64d", &s, &x);
  printf("%I64d\n", solve());
  return 0;
}

### Prompt

In cpp, your task is to solve 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 s, x;
long long solve() {
  long long v = 0;
  for (int i = 0; i < 60; i++) {
    if ((x & (1LL << i)) != 0) {
      v++;
    }
  }
  if (s < x) {
    return 0LL;
  } else if (s == x) {
    return (1LL << v) - 2;
  } else {
    long long d = s - x;
    if ((d & 1) == 1) {
      return 0LL;
    } else {
      for (int i = 0; i < 60; i++) {
        if ((d & (1LL << i)) != 0 && (x & (1LL << (i - 1))) != 0) {
          return 0LL;
        }
      }
      return (1LL << v);
    }
  }
}
int main() {
  scanf("%I64d %I64d", &s, &x);
  printf("%I64d\n", solve());
  return 0;
}

```