#include <bits/stdc++.h>
using namespace std;
long long binpow(long long a, long long b) {
  long long ans = 1, y = a;
  while (b > 0) {
    if (b & 1) {
      ans = (ans * y);
      b--;
      continue;
    }
    y = (y * y);
    b >>= 1;
  }
  return ans;
}
int main() {
  long long s, x, y, ans;
  cin >> s >> x;
  y = s - x;
  if (y < 0) {
    cout << 0 << endl;
    return 0;
  }
  int j;
  j = __builtin_popcountll(x);
  ans = binpow(2, j);
  if (y == 0) {
    ans = max(ans - 2, 1ll * 0);
    cout << ans << endl;
    return 0;
  }
  for (int i = 1; i <= 40; i++) {
    if (!((1ll << i) & y)) continue;
    if ((1ll << (i - 1)) & x) continue;
    y -= (1ll << (i));
  }
  if (y != 0) {
    cout << 0 << endl;
  }
  if (y == 0) {
    cout << ans << endl;
  }
  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 binpow(long long a, long long b) {
  long long ans = 1, y = a;
  while (b > 0) {
    if (b & 1) {
      ans = (ans * y);
      b--;
      continue;
    }
    y = (y * y);
    b >>= 1;
  }
  return ans;
}
int main() {
  long long s, x, y, ans;
  cin >> s >> x;
  y = s - x;
  if (y < 0) {
    cout << 0 << endl;
    return 0;
  }
  int j;
  j = __builtin_popcountll(x);
  ans = binpow(2, j);
  if (y == 0) {
    ans = max(ans - 2, 1ll * 0);
    cout << ans << endl;
    return 0;
  }
  for (int i = 1; i <= 40; i++) {
    if (!((1ll << i) & y)) continue;
    if ((1ll << (i - 1)) & x) continue;
    y -= (1ll << (i));
  }
  if (y != 0) {
    cout << 0 << endl;
  }
  if (y == 0) {
    cout << ans << endl;
  }
  return 0;
}

```