#include <bits/stdc++.h>
using namespace std;
long long a, b;
long long dp[60][2][2][2];
long long solve(int pos, bool carry, bool x, bool y) {
  if (pos > 55) {
    return x && y && (!carry);
  }
  if (dp[pos][carry][x][y] != -1) {
    return dp[pos][carry][x][y];
  }
  long long res = 0;
  for (int i = 0; i < 2; ++i) {
    for (int j = 0; j < 2; ++j) {
      int tmp = i ^ j;
      bool bit = ((b >> pos) & 1);
      if (tmp != bit) {
        continue;
      }
      tmp = (i + j + carry) % 2;
      int xtra = (i + j + carry) / 2;
      bit = ((a >> pos) & 1);
      if (tmp != bit) {
        continue;
      }
      res += solve(pos + 1, xtra, x | (i > 0), y | (j > 0));
    }
  }
  return dp[pos][carry][x][y] = res;
}
int main() {
  memset(dp, -1, sizeof(dp));
  cin >> a >> b;
  cout << solve(0, 0, 0, 0);
}

### Prompt

Please create a solution in cpp to 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 a, b;
long long dp[60][2][2][2];
long long solve(int pos, bool carry, bool x, bool y) {
  if (pos > 55) {
    return x && y && (!carry);
  }
  if (dp[pos][carry][x][y] != -1) {
    return dp[pos][carry][x][y];
  }
  long long res = 0;
  for (int i = 0; i < 2; ++i) {
    for (int j = 0; j < 2; ++j) {
      int tmp = i ^ j;
      bool bit = ((b >> pos) & 1);
      if (tmp != bit) {
        continue;
      }
      tmp = (i + j + carry) % 2;
      int xtra = (i + j + carry) / 2;
      bit = ((a >> pos) & 1);
      if (tmp != bit) {
        continue;
      }
      res += solve(pos + 1, xtra, x | (i > 0), y | (j > 0));
    }
  }
  return dp[pos][carry][x][y] = res;
}
int main() {
  memset(dp, -1, sizeof(dp));
  cin >> a >> b;
  cout << solve(0, 0, 0, 0);
}

```