#include <bits/stdc++.h>
using namespace std;
long long solve(long long s, long long x) {
  vector<long long> xv;
  vector<long long> sv;
  while (x) {
    xv.push_back(x % 2);
    x /= 2;
  }
  while (s) {
    sv.push_back(s % 2);
    s /= 2;
  }
  long long mSz = max(sv.size(), xv.size());
  sv.resize(mSz);
  xv.resize(mSz);
  long long yv = 0;
  long long cnt = 0;
  long long fail = 0;
  long long y = 0;
  for (long long i = mSz - 1; i >= 0; --i) {
    if (yv) {
      if (sv[i] == 1) {
        if (xv[i] == 1) {
          fail = 1;
        } else {
          yv = 1;
          ++y;
        }
      } else {
        if (xv[i] == 1) {
          yv = 1;
          ++cnt;
        } else {
          yv = 0;
          ++y;
        }
      }
    } else {
      if (sv[i] == 1) {
        if (xv[i] == 1) {
          ++cnt;
          yv = 0;
        } else {
          yv = 1;
        }
      } else {
        if (xv[i] == 1) {
          fail = 1;
        } else {
          yv = 0;
        }
      }
    }
  }
  if (fail) return 0;
  long long ans = 1ll << cnt;
  if (yv) return 0;
  if (y == 0 && ans > 1) {
    ans -= 2;
  } else if (y == 0) {
    ans -= 1;
  }
  return ans;
}
int main() {
  long long s, x;
  cin >> s >> x;
  cout << solve(s, x) << '\n';
  return 0;
  for (long long i = 2; i < 11100; ++i) {
    for (long long j = 0; j < 1011; ++j) {
      long long s = rand() % 1000000 + 2;
      long long x = rand() % 1000000;
      long long q = solve(s, x);
      long long w = 0;
      for (long long k = 1; k < s; ++k) {
        long long z = s - k;
        if ((z ^ k) == x) ++w;
      }
      if (q != w) {
        cout << q << ' ' << w << ' ' << s << ' ' << x << '\n';
        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>
using namespace std;
long long solve(long long s, long long x) {
  vector<long long> xv;
  vector<long long> sv;
  while (x) {
    xv.push_back(x % 2);
    x /= 2;
  }
  while (s) {
    sv.push_back(s % 2);
    s /= 2;
  }
  long long mSz = max(sv.size(), xv.size());
  sv.resize(mSz);
  xv.resize(mSz);
  long long yv = 0;
  long long cnt = 0;
  long long fail = 0;
  long long y = 0;
  for (long long i = mSz - 1; i >= 0; --i) {
    if (yv) {
      if (sv[i] == 1) {
        if (xv[i] == 1) {
          fail = 1;
        } else {
          yv = 1;
          ++y;
        }
      } else {
        if (xv[i] == 1) {
          yv = 1;
          ++cnt;
        } else {
          yv = 0;
          ++y;
        }
      }
    } else {
      if (sv[i] == 1) {
        if (xv[i] == 1) {
          ++cnt;
          yv = 0;
        } else {
          yv = 1;
        }
      } else {
        if (xv[i] == 1) {
          fail = 1;
        } else {
          yv = 0;
        }
      }
    }
  }
  if (fail) return 0;
  long long ans = 1ll << cnt;
  if (yv) return 0;
  if (y == 0 && ans > 1) {
    ans -= 2;
  } else if (y == 0) {
    ans -= 1;
  }
  return ans;
}
int main() {
  long long s, x;
  cin >> s >> x;
  cout << solve(s, x) << '\n';
  return 0;
  for (long long i = 2; i < 11100; ++i) {
    for (long long j = 0; j < 1011; ++j) {
      long long s = rand() % 1000000 + 2;
      long long x = rand() % 1000000;
      long long q = solve(s, x);
      long long w = 0;
      for (long long k = 1; k < s; ++k) {
        long long z = s - k;
        if ((z ^ k) == x) ++w;
      }
      if (q != w) {
        cout << q << ' ' << w << ' ' << s << ' ' << x << '\n';
        return 0;
      }
    }
  }
}

```