#include <bits/stdc++.h>
using namespace std;
long long s, x;
long long dp[42][2], a = 1;
long long solve(int i, int neq) {
  if (i < 0) return 1 - neq;
  long long &ret = dp[i][neq];
  if (ret != -1) return ret;
  ret = 0;
  if (neq) {
    if (x & (a << i)) {
      if (s & (a << i))
        ret = 0;
      else
        ret = 2 * solve(i - 1, 1);
      return ret;
    }
    if (s & (a << i))
      ret = solve(i - 1, 1);
    else
      ret = solve(i - 1, 0);
  } else {
    if (s & (a << i)) {
      if (x & (a << i)) {
        ret = 2 * solve(i - 1, 0);
      } else {
        ret = solve(i - 1, 1);
      }
    } else {
      if (x & (a << i)) {
        ret = 0;
      } else
        ret = solve(i - 1, 0);
    }
  }
  return ret;
}
int main() {
  cin >> s >> x;
  memset(dp, -1, sizeof dp);
  long long ans = solve(41, 0);
  if (x == s) ans -= 2;
  cout << ans;
}

### 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;
long long s, x;
long long dp[42][2], a = 1;
long long solve(int i, int neq) {
  if (i < 0) return 1 - neq;
  long long &ret = dp[i][neq];
  if (ret != -1) return ret;
  ret = 0;
  if (neq) {
    if (x & (a << i)) {
      if (s & (a << i))
        ret = 0;
      else
        ret = 2 * solve(i - 1, 1);
      return ret;
    }
    if (s & (a << i))
      ret = solve(i - 1, 1);
    else
      ret = solve(i - 1, 0);
  } else {
    if (s & (a << i)) {
      if (x & (a << i)) {
        ret = 2 * solve(i - 1, 0);
      } else {
        ret = solve(i - 1, 1);
      }
    } else {
      if (x & (a << i)) {
        ret = 0;
      } else
        ret = solve(i - 1, 0);
    }
  }
  return ret;
}
int main() {
  cin >> s >> x;
  memset(dp, -1, sizeof dp);
  long long ans = solve(41, 0);
  if (x == s) ans -= 2;
  cout << ans;
}

```