#include <bits/stdc++.h>
using namespace std;
long long s, x, a[1000 + 1], n, kq = 1;
void doc() { cin >> s >> x; }
bool cmp(long long i, long long j) { return i > j; }
void xl() {
  s -= x;
  long long bit = 0, gt = 1;
  for (int i = 1; i <= 60; i++) {
    if (x % 2 == 0) {
      a[++n] = 2 * gt;
    } else
      kq *= 2;
    bit++;
    gt *= 2;
    x /= 2;
  }
  sort(a + 1, a + n + 1, cmp);
  long long demd = 0;
  for (int i = 1; i <= n; i++) {
    if (s >= a[i]) s -= a[i], demd++;
  }
  if (demd == 0) kq -= 2;
  if (s == 0)
    cout << kq;
  else
    cout << 0;
}
int main() {
  cin.tie(NULL);
  ios_base::sync_with_stdio(false);
  doc();
  xl();
}

### Prompt

Please formulate 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;
long long s, x, a[1000 + 1], n, kq = 1;
void doc() { cin >> s >> x; }
bool cmp(long long i, long long j) { return i > j; }
void xl() {
  s -= x;
  long long bit = 0, gt = 1;
  for (int i = 1; i <= 60; i++) {
    if (x % 2 == 0) {
      a[++n] = 2 * gt;
    } else
      kq *= 2;
    bit++;
    gt *= 2;
    x /= 2;
  }
  sort(a + 1, a + n + 1, cmp);
  long long demd = 0;
  for (int i = 1; i <= n; i++) {
    if (s >= a[i]) s -= a[i], demd++;
  }
  if (demd == 0) kq -= 2;
  if (s == 0)
    cout << kq;
  else
    cout << 0;
}
int main() {
  cin.tie(NULL);
  ios_base::sync_with_stdio(false);
  doc();
  xl();
}

```