#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
const double PI = acos(-1);
long long s, x;
vector<int> S, X;
vector<int> num(long long s) {
  vector<int> res;
  while (s > 0) {
    res.push_back(s % 2);
    s /= 2;
  }
  return res;
}
vector<int> add(vector<int> p, int len) {
  int temp = p.size();
  for (int i = temp; i < len; i++) p.push_back(0);
  p.push_back(0);
  return p;
}
int main() {
  cin >> s >> x;
  S = num(s);
  X = num(x);
  S = add(S, max(((int)(S).size()), ((int)(X).size())));
  X = add(X, max(((int)(S).size()), ((int)(X).size())));
  bool ok = true;
  int c = 0;
  long long ans = 1;
  for (int i = 0; i < ((int)(S).size()) - 1; i++) {
    if (X[i] == 1) {
      if (S[i] == c) {
        ok = false;
        break;
      }
      ans = ans * 2;
    } else {
      if (S[i] != c) {
        ok = false;
        break;
      } else {
        if (S[i + 1] == X[i + 1]) {
          c = 0;
        } else {
          c = 1;
        }
      }
    }
  }
  if (c == 1) ok = false;
  if (s == x) ans -= 2;
  if (ok)
    cout << ans << endl;
  else
    cout << "0" << endl;
  return 0;
}

### Prompt

Generate 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;
const long long mod = 1e9 + 7;
const double PI = acos(-1);
long long s, x;
vector<int> S, X;
vector<int> num(long long s) {
  vector<int> res;
  while (s > 0) {
    res.push_back(s % 2);
    s /= 2;
  }
  return res;
}
vector<int> add(vector<int> p, int len) {
  int temp = p.size();
  for (int i = temp; i < len; i++) p.push_back(0);
  p.push_back(0);
  return p;
}
int main() {
  cin >> s >> x;
  S = num(s);
  X = num(x);
  S = add(S, max(((int)(S).size()), ((int)(X).size())));
  X = add(X, max(((int)(S).size()), ((int)(X).size())));
  bool ok = true;
  int c = 0;
  long long ans = 1;
  for (int i = 0; i < ((int)(S).size()) - 1; i++) {
    if (X[i] == 1) {
      if (S[i] == c) {
        ok = false;
        break;
      }
      ans = ans * 2;
    } else {
      if (S[i] != c) {
        ok = false;
        break;
      } else {
        if (S[i + 1] == X[i + 1]) {
          c = 0;
        } else {
          c = 1;
        }
      }
    }
  }
  if (c == 1) ok = false;
  if (s == x) ans -= 2;
  if (ok)
    cout << ans << endl;
  else
    cout << "0" << endl;
  return 0;
}

```