#include <bits/stdc++.h>
std::vector<int64_t> bin(int64_t n) {
  std::vector<int64_t> result;
  while (n > 0) {
    result.push_back(n % 2);
    n /= 2;
  }
  return result;
}
int main() {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(nullptr);
  int64_t s, x;
  std::cin >> s >> x;
  auto s_v = bin(s);
  auto x_v = bin(x);
  int size = std::max(s_v.size(), x_v.size());
  std::vector<std::vector<int64_t>> dp(100, std::vector<int64_t>(2));
  dp[0][0] = (s_v[0] == 0 && (x_v.empty() || x_v[0] == 0)) ? 1 : 2;
  dp[0][1] = (s_v[0] == 0 && (x_v.empty() || x_v[0] == 0)) ? 1 : 0;
  for (int i = 1; i < size; ++i) {
    if (i < x_v.size() && x_v[i] == 1) {
      if (i >= s_v.size() || s_v[i] == 0) {
        dp[i][1] = 2 * dp[i - 1][1];
        dp[i][0] = 0;
      } else {
        dp[i][0] = 2 * dp[i - 1][0];
        dp[i][1] = 0;
      }
    } else {
      if (i >= s_v.size() || s_v[i] == 0) {
        dp[i][0] = dp[i - 1][0];
        dp[i][1] = dp[i - 1][0];
      } else {
        dp[i][1] = dp[i - 1][1];
        dp[i][0] = dp[i - 1][1];
      }
    }
  }
  if (x == s) {
    std::cout << dp[size - 1][0] - 2 << std::endl;
  } else {
    std::cout << dp[size - 1][0] << std::endl;
  }
  return 0;
}

### Prompt

Your task is to create a CPP solution 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>
std::vector<int64_t> bin(int64_t n) {
  std::vector<int64_t> result;
  while (n > 0) {
    result.push_back(n % 2);
    n /= 2;
  }
  return result;
}
int main() {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(nullptr);
  int64_t s, x;
  std::cin >> s >> x;
  auto s_v = bin(s);
  auto x_v = bin(x);
  int size = std::max(s_v.size(), x_v.size());
  std::vector<std::vector<int64_t>> dp(100, std::vector<int64_t>(2));
  dp[0][0] = (s_v[0] == 0 && (x_v.empty() || x_v[0] == 0)) ? 1 : 2;
  dp[0][1] = (s_v[0] == 0 && (x_v.empty() || x_v[0] == 0)) ? 1 : 0;
  for (int i = 1; i < size; ++i) {
    if (i < x_v.size() && x_v[i] == 1) {
      if (i >= s_v.size() || s_v[i] == 0) {
        dp[i][1] = 2 * dp[i - 1][1];
        dp[i][0] = 0;
      } else {
        dp[i][0] = 2 * dp[i - 1][0];
        dp[i][1] = 0;
      }
    } else {
      if (i >= s_v.size() || s_v[i] == 0) {
        dp[i][0] = dp[i - 1][0];
        dp[i][1] = dp[i - 1][0];
      } else {
        dp[i][1] = dp[i - 1][1];
        dp[i][0] = dp[i - 1][1];
      }
    }
  }
  if (x == s) {
    std::cout << dp[size - 1][0] - 2 << std::endl;
  } else {
    std::cout << dp[size - 1][0] << std::endl;
  }
  return 0;
}

```