#include <bits/stdc++.h>
using namespace std;
const int N = 60;
int a[N], b[N];
long long dp[N][2];
long long f(int p, int c) {
  if (p == N) return c == 0;
  long long &ret = dp[p][c];
  if (ret != -1) return ret;
  ret = 0;
  for (int i = 0; i < 2; ++i)
    for (int j = 0; j < 2; ++j)
      if ((i ^ j ^ c) == a[p] && (i ^ j) == b[p])
        ret += f(p + 1, i + j + c > 1);
  return ret;
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  long long s, x;
  cin >> s >> x;
  for (int i = 0; i < N; ++i) {
    a[i] = (s >> i & 1);
    b[i] = (x >> i & 1);
  }
  memset(dp, -1, sizeof dp);
  cout << f(0, 0) - (s == x) * 2 << "\n";
  return 0;
}

### Prompt

Develop a solution in Cpp 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 = 60;
int a[N], b[N];
long long dp[N][2];
long long f(int p, int c) {
  if (p == N) return c == 0;
  long long &ret = dp[p][c];
  if (ret != -1) return ret;
  ret = 0;
  for (int i = 0; i < 2; ++i)
    for (int j = 0; j < 2; ++j)
      if ((i ^ j ^ c) == a[p] && (i ^ j) == b[p])
        ret += f(p + 1, i + j + c > 1);
  return ret;
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  long long s, x;
  cin >> s >> x;
  for (int i = 0; i < N; ++i) {
    a[i] = (s >> i & 1);
    b[i] = (x >> i & 1);
  }
  memset(dp, -1, sizeof dp);
  cout << f(0, 0) - (s == x) * 2 << "\n";
  return 0;
}

```