#include <bits/stdc++.h>
using namespace std;
vector<long long> pp(long long k) {
  vector<long long> v;
  while (k) {
    v.push_back(k % 2);
    k /= 2;
  }
  return v;
}
signed main() {
  long long n, m;
  cin >> n >> m;
  vector<long long> vn = pp(n);
  vector<long long> vm = pp(m);
  long long ss = 0;
  while (vm.size() < 50) vm.push_back(0);
  while (vn.size() < 50) vn.push_back(0);
  long long cnt = 0;
  for (long long i = 0; i < vm.size(); ++i) {
    if (vm[i] == 1) {
      ss += 1LL << i;
      ++cnt;
    }
  }
  n -= ss;
  if (n < 0) return 0 * printf("0");
  vector<long long> kk = pp(n);
  for (long long i = 0; i < kk.size(); ++i) {
    if (kk[i] == 1) {
      if (i != 0 && vm[i - 1] == 0) {
        continue;
      } else {
        return 0 * printf("0");
      }
    }
  }
  if (n == 0) {
    cout << max(0LL, (1LL << cnt) - 2) << endl;
  } else {
    cout << (1LL << cnt) << endl;
  }
}

### Prompt

Develop a solution in cpp to the problem described below:

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;
vector<long long> pp(long long k) {
  vector<long long> v;
  while (k) {
    v.push_back(k % 2);
    k /= 2;
  }
  return v;
}
signed main() {
  long long n, m;
  cin >> n >> m;
  vector<long long> vn = pp(n);
  vector<long long> vm = pp(m);
  long long ss = 0;
  while (vm.size() < 50) vm.push_back(0);
  while (vn.size() < 50) vn.push_back(0);
  long long cnt = 0;
  for (long long i = 0; i < vm.size(); ++i) {
    if (vm[i] == 1) {
      ss += 1LL << i;
      ++cnt;
    }
  }
  n -= ss;
  if (n < 0) return 0 * printf("0");
  vector<long long> kk = pp(n);
  for (long long i = 0; i < kk.size(); ++i) {
    if (kk[i] == 1) {
      if (i != 0 && vm[i - 1] == 0) {
        continue;
      } else {
        return 0 * printf("0");
      }
    }
  }
  if (n == 0) {
    cout << max(0LL, (1LL << cnt) - 2) << endl;
  } else {
    cout << (1LL << cnt) << endl;
  }
}

```