#include <bits/stdc++.h>
using namespace std;
long long bits[60];
long long ans = 0;
long long s, t;
void dfs(long long a, long long b, int pos, long long c) {
  if (a + b == s && (a ^ b) == t) {
    ans += c;
    return;
  } else if (a + b > s) {
    return;
  }
  if ((t & bits[pos]) == bits[pos]) {
    if ((a + bits[pos] + b) % bits[pos + 1] == s % bits[pos + 1])
      dfs(a + bits[pos], b, pos + 1, c * 2);
  } else {
    if ((a + b) % bits[pos + 1] == s % bits[pos + 1]) {
      dfs(a, b, pos + 1, c);
      dfs(a + bits[pos], b + bits[pos], pos + 1, c);
    }
  }
}
int main() {
  bits[0] = 1LL;
  for (int i = 1; i < 60; i++) {
    bits[i] = bits[i - 1] << 1;
  }
  cin >> s >> t;
  if ((s & 1) != (t & 1)) {
    cout << 0 << endl;
    return 0;
  } else {
    if (t % bits[1] == 1) {
      dfs(0, 1, 1, 2);
    } else {
      dfs(0, 0, 1, 1);
      dfs(1, 1, 1, 1);
    }
  }
  if (s == t) {
    ans -= 2;
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Please create a solution in cpp 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 bits[60];
long long ans = 0;
long long s, t;
void dfs(long long a, long long b, int pos, long long c) {
  if (a + b == s && (a ^ b) == t) {
    ans += c;
    return;
  } else if (a + b > s) {
    return;
  }
  if ((t & bits[pos]) == bits[pos]) {
    if ((a + bits[pos] + b) % bits[pos + 1] == s % bits[pos + 1])
      dfs(a + bits[pos], b, pos + 1, c * 2);
  } else {
    if ((a + b) % bits[pos + 1] == s % bits[pos + 1]) {
      dfs(a, b, pos + 1, c);
      dfs(a + bits[pos], b + bits[pos], pos + 1, c);
    }
  }
}
int main() {
  bits[0] = 1LL;
  for (int i = 1; i < 60; i++) {
    bits[i] = bits[i - 1] << 1;
  }
  cin >> s >> t;
  if ((s & 1) != (t & 1)) {
    cout << 0 << endl;
    return 0;
  } else {
    if (t % bits[1] == 1) {
      dfs(0, 1, 1, 2);
    } else {
      dfs(0, 0, 1, 1);
      dfs(1, 1, 1, 1);
    }
  }
  if (s == t) {
    ans -= 2;
  }
  cout << ans << endl;
  return 0;
}

```