#include <bits/stdc++.h>
using namespace std;
long long dp[50][2];
long long second, X;
void give(int n, long long c, long long x, long long s) {
  if (x > 1) x = 1;
  if (s > 1) s = 1;
  for (int i = 0; i < 2; i++) {
    for (int j = 0; j < 2; j++) {
      if ((i ^ j) == x and (c + i + j) % 2 == s) {
        dp[n + 1][(c + i + j) / 2] += dp[n][c];
      }
    }
  }
}
inline void ReadInput(void) {
  scanf("%lld", &second);
  scanf("%lld", &X);
}
inline void solve(void) {
  dp[0][0] = 1;
  for (int i = 0; i < 50; i++) {
    give(i, 0, X & (1LL << i), second & (1LL << i));
    give(i, 1, X & (1LL << i), second & (1LL << i));
  }
  if (second == X) dp[49][0] -= 2;
  cout << dp[49][0] << endl;
}
inline void Refresh(void) {}
int main() {
  ReadInput();
  solve();
  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 dp[50][2];
long long second, X;
void give(int n, long long c, long long x, long long s) {
  if (x > 1) x = 1;
  if (s > 1) s = 1;
  for (int i = 0; i < 2; i++) {
    for (int j = 0; j < 2; j++) {
      if ((i ^ j) == x and (c + i + j) % 2 == s) {
        dp[n + 1][(c + i + j) / 2] += dp[n][c];
      }
    }
  }
}
inline void ReadInput(void) {
  scanf("%lld", &second);
  scanf("%lld", &X);
}
inline void solve(void) {
  dp[0][0] = 1;
  for (int i = 0; i < 50; i++) {
    give(i, 0, X & (1LL << i), second & (1LL << i));
    give(i, 1, X & (1LL << i), second & (1LL << i));
  }
  if (second == X) dp[49][0] -= 2;
  cout << dp[49][0] << endl;
}
inline void Refresh(void) {}
int main() {
  ReadInput();
  solve();
  return 0;
}

```