#include <bits/stdc++.h>
using namespace std;
long long int dp[50][2];
long long int sum;
long long int xo;
int n;
long long int solve(int bit, int carry) {
  if (bit == n) return (carry) ? 0LL : 1LL;
  long long int bitSum = (sum >> bit) & 1;
  long long int bitXor = (xo >> bit) & 1;
  long long int &res = dp[bit][carry];
  if (res != -1) return res;
  res = 0;
  if (bitSum == 0 and bitXor == 0) {
    if (!carry) {
      res = solve(bit + 1, 1) + solve(bit + 1, 0);
    }
  } else if (bitSum == 1 and bitXor == 1) {
    if (!carry) {
      res = 2LL * solve(bit + 1, 0);
    }
  } else if (bitSum == 0 and bitXor == 1) {
    if (carry) {
      res = 2LL * solve(bit + 1, 1);
    }
  } else {
    if (carry) {
      res = solve(bit + 1, 1) + solve(bit + 1, 0);
    }
  }
  return res;
}
int main(void) {
  cin >> sum >> xo;
  long long int m = max(sum, xo);
  for (int i = 0; i < 50; ++i) {
    if ((m >> i) & 1) {
      n = i + 1;
    }
  }
  memset(dp, -1, sizeof dp);
  long long int res = solve(0, 0);
  if (sum == xo) res -= 2;
  cout << res << endl;
  return 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 int dp[50][2];
long long int sum;
long long int xo;
int n;
long long int solve(int bit, int carry) {
  if (bit == n) return (carry) ? 0LL : 1LL;
  long long int bitSum = (sum >> bit) & 1;
  long long int bitXor = (xo >> bit) & 1;
  long long int &res = dp[bit][carry];
  if (res != -1) return res;
  res = 0;
  if (bitSum == 0 and bitXor == 0) {
    if (!carry) {
      res = solve(bit + 1, 1) + solve(bit + 1, 0);
    }
  } else if (bitSum == 1 and bitXor == 1) {
    if (!carry) {
      res = 2LL * solve(bit + 1, 0);
    }
  } else if (bitSum == 0 and bitXor == 1) {
    if (carry) {
      res = 2LL * solve(bit + 1, 1);
    }
  } else {
    if (carry) {
      res = solve(bit + 1, 1) + solve(bit + 1, 0);
    }
  }
  return res;
}
int main(void) {
  cin >> sum >> xo;
  long long int m = max(sum, xo);
  for (int i = 0; i < 50; ++i) {
    if ((m >> i) & 1) {
      n = i + 1;
    }
  }
  memset(dp, -1, sizeof dp);
  long long int res = solve(0, 0);
  if (sum == xo) res -= 2;
  cout << res << endl;
  return 0;
}

```