#include <bits/stdc++.h>
using namespace std;
const int N = 41;
long long s, xo;
long long memo[N][2];
bool get(long long s, int i) { return (s >> i) & 1; }
long long solve(int i, bool carry) {
  if (i == N) return !carry;
  auto &ret = memo[i][carry];
  if (~ret) return ret;
  ret = 0;
  for (int x = 0; x < 2; x++) {
    for (int y = 0; y < 2; y++) {
      int r = carry + x + y;
      if (x ^ y == get(xo, i) && get(r, 0) == get(s, i)) {
        ret += solve(i + 1, get(r, 1));
      }
    }
  }
  return ret;
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> s >> xo;
  memset(memo, -1, sizeof memo);
  long long ans = solve(0, 0);
  if (s == xo) ans -= 2;
  cout << ans;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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;
const int N = 41;
long long s, xo;
long long memo[N][2];
bool get(long long s, int i) { return (s >> i) & 1; }
long long solve(int i, bool carry) {
  if (i == N) return !carry;
  auto &ret = memo[i][carry];
  if (~ret) return ret;
  ret = 0;
  for (int x = 0; x < 2; x++) {
    for (int y = 0; y < 2; y++) {
      int r = carry + x + y;
      if (x ^ y == get(xo, i) && get(r, 0) == get(s, i)) {
        ret += solve(i + 1, get(r, 1));
      }
    }
  }
  return ret;
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> s >> xo;
  memset(memo, -1, sizeof memo);
  long long ans = solve(0, 0);
  if (s == xo) ans -= 2;
  cout << ans;
}

```