#include <bits/stdc++.h>
using namespace std;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  long long s, x, sdash;
  cin >> s >> x;
  sdash = s - x;
  vector<long long> sbits(40, 0), xbits(40, 0), sdashbits(40, 0);
  for (long long i = 0; i < 40; i++) {
    if (s & (1LL << i)) {
      sbits[i] = 1;
    }
    if (x & (1LL << i)) {
      xbits[i] = 1;
    }
    if (sdash & (1LL << i)) {
      sdashbits[i] = 1;
    }
  }
  if (sdashbits[0] != 0) {
    cout << "0\n";
    return 0;
  }
  for (long long i = 1; i < 40; i++) {
    if (sdashbits[i] == 1) {
      if (xbits[i - 1] != 0) {
        cout << "0\n";
        return 0;
      }
    }
  }
  long long cnt = 0;
  for (long long i = 0; i < 40; i++) {
    cnt += xbits[i];
  }
  long long ans = (1LL << cnt);
  if (s == x) {
    ans -= 2;
  }
  cout << ans << "\n";
}

### Prompt

Your task is to create 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;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  long long s, x, sdash;
  cin >> s >> x;
  sdash = s - x;
  vector<long long> sbits(40, 0), xbits(40, 0), sdashbits(40, 0);
  for (long long i = 0; i < 40; i++) {
    if (s & (1LL << i)) {
      sbits[i] = 1;
    }
    if (x & (1LL << i)) {
      xbits[i] = 1;
    }
    if (sdash & (1LL << i)) {
      sdashbits[i] = 1;
    }
  }
  if (sdashbits[0] != 0) {
    cout << "0\n";
    return 0;
  }
  for (long long i = 1; i < 40; i++) {
    if (sdashbits[i] == 1) {
      if (xbits[i - 1] != 0) {
        cout << "0\n";
        return 0;
      }
    }
  }
  long long cnt = 0;
  for (long long i = 0; i < 40; i++) {
    cnt += xbits[i];
  }
  long long ans = (1LL << cnt);
  if (s == x) {
    ans -= 2;
  }
  cout << ans << "\n";
}

```