#include <bits/stdc++.h>
using namespace std;
long long d[100][2];
int main() {
  ios_base::sync_with_stdio(false);
  long long x, s;
  cin >> s >> x;
  if (x > s) {
    cout << 0 << endl;
    return 0;
  }
  long long z = s;
  int n = 0;
  while (z > 0) {
    z /= 2;
    n++;
  }
  d[0][0] = 1ll;
  for (int i = 1; i <= n; i++) {
    if (((s >> (i - 1)) % 2 == 0) && ((x >> (i - 1)) % 2 == 0)) {
      d[i][0] = d[i - 1][0];
      d[i][1] = d[i - 1][0];
    }
    if (((s >> (i - 1)) % 2 == 1) && ((x >> (i - 1)) % 2 == 0)) {
      d[i][0] = d[i - 1][1];
      d[i][1] = d[i - 1][1];
    }
    if (((s >> (i - 1)) % 2 == 0) && ((x >> (i - 1)) % 2 == 1)) {
      d[i][0] = 0ll;
      d[i][1] = d[i - 1][1] * 2;
    }
    if (((s >> (i - 1)) % 2 == 1) && ((x >> (i - 1)) % 2 == 1)) {
      d[i][0] = d[i - 1][0] * 2;
      d[i][1] = 0ll;
    }
  }
  if (x == s) d[n][0] -= 2ll;
  cout << d[n][0] << 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 d[100][2];
int main() {
  ios_base::sync_with_stdio(false);
  long long x, s;
  cin >> s >> x;
  if (x > s) {
    cout << 0 << endl;
    return 0;
  }
  long long z = s;
  int n = 0;
  while (z > 0) {
    z /= 2;
    n++;
  }
  d[0][0] = 1ll;
  for (int i = 1; i <= n; i++) {
    if (((s >> (i - 1)) % 2 == 0) && ((x >> (i - 1)) % 2 == 0)) {
      d[i][0] = d[i - 1][0];
      d[i][1] = d[i - 1][0];
    }
    if (((s >> (i - 1)) % 2 == 1) && ((x >> (i - 1)) % 2 == 0)) {
      d[i][0] = d[i - 1][1];
      d[i][1] = d[i - 1][1];
    }
    if (((s >> (i - 1)) % 2 == 0) && ((x >> (i - 1)) % 2 == 1)) {
      d[i][0] = 0ll;
      d[i][1] = d[i - 1][1] * 2;
    }
    if (((s >> (i - 1)) % 2 == 1) && ((x >> (i - 1)) % 2 == 1)) {
      d[i][0] = d[i - 1][0] * 2;
      d[i][1] = 0ll;
    }
  }
  if (x == s) d[n][0] -= 2ll;
  cout << d[n][0] << endl;
  return 0;
}

```