#include <bits/stdc++.h>
using namespace std;
long long s, x;
long long dp[66][2];
long long solve(int pos, int carry) {
  if (pos == 50) {
    return (carry == 0);
  }
  if (dp[pos][carry] != -1) return dp[pos][carry];
  long long answer = 0;
  for (int b1 = 0; b1 <= 1; b1++)
    for (int b2 = 0; b2 <= 1; b2++) {
      if ((b1 ^ b2) != ((x >> pos) & 1)) continue;
      if ((b1 + b2 + carry) % 2 != ((s >> pos) & 1)) continue;
      answer += solve(pos + 1, (b1 + b2 + carry) / 2);
    }
  return dp[pos][carry] = answer;
}
long long solvebrute() {
  long long ct = 0;
  for (int a = 1; a <= s; a++)
    for (int b = 1; b <= s - a; b++)
      if (a + b == s && (a ^ b) == x) ct++;
  return ct;
}
int main() {
  cin >> s >> x;
  for (int i = 0; i <= 60; i++) dp[i][0] = dp[i][1] = -1;
  long long answer = solve(0, 0);
  if (s == x) {
    answer -= 2;
  }
  cout << answer << endl;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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[66][2];
long long solve(int pos, int carry) {
  if (pos == 50) {
    return (carry == 0);
  }
  if (dp[pos][carry] != -1) return dp[pos][carry];
  long long answer = 0;
  for (int b1 = 0; b1 <= 1; b1++)
    for (int b2 = 0; b2 <= 1; b2++) {
      if ((b1 ^ b2) != ((x >> pos) & 1)) continue;
      if ((b1 + b2 + carry) % 2 != ((s >> pos) & 1)) continue;
      answer += solve(pos + 1, (b1 + b2 + carry) / 2);
    }
  return dp[pos][carry] = answer;
}
long long solvebrute() {
  long long ct = 0;
  for (int a = 1; a <= s; a++)
    for (int b = 1; b <= s - a; b++)
      if (a + b == s && (a ^ b) == x) ct++;
  return ct;
}
int main() {
  cin >> s >> x;
  for (int i = 0; i <= 60; i++) dp[i][0] = dp[i][1] = -1;
  long long answer = solve(0, 0);
  if (s == x) {
    answer -= 2;
  }
  cout << answer << endl;
}

```