#include <bits/stdc++.h>
using namespace std;
long long s, x;
long long dp[42][2];
long long rec(int digit, int carry) {
  if (dp[digit][carry] >= 0) return dp[digit][carry];
  long long ret = 0;
  if (digit == 41) {
    if (carry)
      return 0;
    else
      return 1;
  }
  if ((x >> digit) & 1) {
    if ((s >> digit) & 1) {
      if (!carry) ret += 2 * rec(digit + 1, 0);
    } else {
      if (carry) ret += 2 * rec(digit + 1, 1);
    }
  } else {
    if ((s >> digit) & 1) {
      if (carry) {
        ret += rec(digit + 1, 0) + rec(digit + 1, 1);
      }
    } else {
      if (!carry) {
        ret += rec(digit + 1, 0) + rec(digit + 1, 1);
      }
    }
  }
  return dp[digit][carry] = ret;
}
int main(int argc, char const *argv[]) {
  cin >> s >> x;
  memset(dp, -1, sizeof(dp));
  long long ans = rec(0, 0);
  if (s == x) {
    ans -= 2;
  }
  std::cout << ans << std::endl;
  return 0;
}

### 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[42][2];
long long rec(int digit, int carry) {
  if (dp[digit][carry] >= 0) return dp[digit][carry];
  long long ret = 0;
  if (digit == 41) {
    if (carry)
      return 0;
    else
      return 1;
  }
  if ((x >> digit) & 1) {
    if ((s >> digit) & 1) {
      if (!carry) ret += 2 * rec(digit + 1, 0);
    } else {
      if (carry) ret += 2 * rec(digit + 1, 1);
    }
  } else {
    if ((s >> digit) & 1) {
      if (carry) {
        ret += rec(digit + 1, 0) + rec(digit + 1, 1);
      }
    } else {
      if (!carry) {
        ret += rec(digit + 1, 0) + rec(digit + 1, 1);
      }
    }
  }
  return dp[digit][carry] = ret;
}
int main(int argc, char const *argv[]) {
  cin >> s >> x;
  memset(dp, -1, sizeof(dp));
  long long ans = rec(0, 0);
  if (s == x) {
    ans -= 2;
  }
  std::cout << ans << std::endl;
  return 0;
}

```